Projectile Motion in Physics - Examples with answers

a) For a given altitude, the flight time is:

$$h=\frac{1}{2}gt^2$$

$$t=\sqrt{\frac{2h}{g}}$$

Substituting the values ​​$latex h= 100\hspace{1mm}m$ and $latex g=9.8\hspace{1mm}\dfrac{m}{s^2}$, it results:

$$t=\sqrt{\frac{2\times 100\hspace{1mm}m }{9.8\hspace{1mm}\dfrac{m}{s^2}}}=4.5\hspace{1mm}s$$

b) The equation for the velocity vector as a function of time is:

$$\vec{v}(t)=v_0\hat{ i} - gt \hat{j}$$

The negative direction is downward, and the velocity components are:

$latex v_{x}=v_0=40\hspace{1mm}\dfrac{m}{s}$

$latex v_y=gt=9.8\hspace{1mm}\dfrac{m}{s^2}\times 4.5\hspace{1mm}s=44.1\hspace{1mm}\dfrac{m}{s}$

Therefore:

$$\vec{v}(t)=40\hat{ i} - 44.1 \hat{j}\hspace{1mm}\frac{m}{s}$$

The magnitude of the final velocity is:

$$v=\sqrt{40^2+(-44.1)^2}\hspace{1mm}\frac{m}{s}=59.5\hspace{1mm}\frac{m}{s}$$

c) The horizontal range is calculated by:

$latex x_{max}=v_x\cdot t_{flight}$

Substituting values:

$$ x_{max}=40\hspace{1mm}\frac{m}{s}\times 4.5\hspace{1mm}s=180\hspace{1mm}m$$

a) Let $latex y_0$ be the height from which the key ring starts, which did not descend completely, since it lacked 1.2 m, since it was caught by the friend. The vertical position is set as:

$$y=y_0-\frac{1}{2}gt^2$$

Substituting values:

$$1.2=y_0-\frac{1}{2}gt^2\Rightarrow y_0=\left(1.2+\frac{1}{2}\times 9.8\times 0.8^2\hspace{1mm}\right)m=4.34\hspace{1mm}m$$

b) First, the velocity components are calculated:

Horizontal component

$$v_x=\frac{x}{t}=\frac{4.8\hspace{1mm}m}{0.8\hspace{1mm}s}=6.0\hspace{1mm}\frac{m}{s}$$

Vertical component

$$v_y=gt=9.8\hspace{1mm}\frac{m}{s}\times 0.8\hspace{1mm}s =7.8\hspace{1mm}\frac{m}{s}$$

The velocity vector is:

$$\vec{v}(t)=6.0\hat{ i} - 7.8 \hat{j}\hspace{1mm}\frac{m}{s}$$

a) From the equation for the horizontal range, the velocity $latex v_0$ with which the toy was initially projected can be calculated:

$$x_{max}=v_{0x}t=v_0t$$

However, we first need to know the time it took to fall. This can be determined from the height of the table:

$$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 74.0\hspace{1mm}cm}{981\hspace{1mm}\dfrac {cm}{s^2}}}=0.39\hspace{1mm}s$$

Knowing the time and the horizontal range of the projectile, which is 97.0 cm according to the statement, the horizontal velocity with which the toy was projected is:

$$x_{max}=v_{0x}t=v_0t\Rightarrow v_{0}=\frac{x_{max}}{t}=\frac{97.0\hspace{1mm}cm}{0.39\hspace{1mm}s}=248.7\hspace{1mm}\frac{cm}{s}$$

It remains constant throughout the motion.

b) The vector equation for the velocity at any instant of motion is:

$$\vec{v}(t)=v_0\hat{ i} - gt \hat{j}$$

It is assigned the negative downward direction.

The horizontal component of the velocity remains constant throughout the motion, and we only need to know the vertical component of the velocity upon reaching the ground.

Calculation of the vertical component of the velocity.

$$v_y=gt\Rightarrow v_y=981\hspace{1mm} \frac{cm}{s^2}\times 0.39\hspace{1mm}s=382.6\hspace{1mm} \frac{cm}{s}$$

Calculation of the final velocity vector.

The results obtained are substituted into the equation for the velocity:

$$\vec{v}(t)=248.7\hat{ i} - 382.6 \hat{j}\hspace{1mm} \frac{cm}{s}$$

The magnitude of the velocity on reaching the ground is:

$$v=\sqrt{248.7^2+(-382.6)^2}\hspace{1mm}\frac{cm}{s}=456\hspace{1mm}\frac{cm}{s}$$

c) The angle formed by the velocity vector with the positive horizontal axis is:

$$\theta =\arctan \frac{v_y}{v_x}=\frac{-382.6}{248.7}=-57º$$

This means that the velocity forms an angle of approximately 57º below the horizontal when the toy impacts the floor.

a) When the object reaches its maximum height, the vertical velocity $latex v_y$ cancels out, and then reverses its direction. In this case:

$$v_y=v_{0y}-gt_{max}=0$$

Where $latex t_{max}$ is the time it takes for the projectile to reach that maximum height. Therefore:

$$t_{max}=\frac{v_{0y}}{g}=\frac{50\hspace{1mm}\dfrac{m}{s}\times sin 30º}{9.8\hspace{1mm}\dfrac{m}{s^2}}=2.6\hspace{1mm}s$$

b) The maximum height is calculated by:

$$y_{max}=\frac{1}{2}gt_{max}^2=\frac{1}{2}\times 9.8\hspace{1mm}\dfrac{m}{s^2}\times (2.6\hspace{1mm}s)^2=33.1\hspace{1mm}m$$

c) Since the projectile falls further forward and at the same level as the ground, the maximum time is:

$$t_{flight}=2\cdot t_{max}=2\times2.6\hspace{1mm}s=5.2\hspace{1mm}s$$

d) The maximum horizontal range, $latex x_{max}$, is given by:

$$x_{max}=v_{0x}\cdot t_{flight}=v_{0}\cos 30º\cdot t_{flight}$$

$$x_{max}=50\hspace{1mm}\frac{m}{s}\times cos 30º\times 5.2\hspace{1mm}s=225.2\hspace{1mm}m$$

a) In vector form, the position as a function of time $latex \vec {r}(t)$ is given by:

$$\vec {r}(t)=(x_0+v_{0x}t)\hat i +\left[y_0+v_{0y}t-\frac{1}{2}gt^2\right]\hat j$$

In the vertical, the negative direction is chosen downward. Choosing $latex x_0=0$ and $latex y_0=0$, the position remains:

$$\vec {r}(t)=v_{0x}t\hspace{1mm}\hat i +\left[v_{0y}t-\frac{1}{2}gt^2\right]\hat j$$

Then, for the following calculations, it is necessary to determine the components of the initial velocity:

$latex v_{0x}=v_0\cos 25º = 200\times \cos 25º\hspace{1mm}\dfrac{m}{s}=181.3\hspace{1mm}\dfrac{m}{s}$

$latex v_{0y}=v_0\cos 25º = 200\times \sin25º\hspace{1mm}\dfrac{m}{s}=84.5\hspace{1mm}\dfrac{m}{s}$

The flight time is obtained from the vertical component of the position equation, since when falling into the sea, the projectile must be in the position $latex y = -150\hspace{1mm}m$, then:

$$v_{0y}t-\frac{1}{2}gt^2=-150$$

Substituting the known values, we obtain a second-degree equation for time:

$$84.5t-4.9t^2=-150$$

$$4.9t^2-84.5t-150=0$$

Whose solutions are:

$latex t_1=-1.6$

$latex t_2=18.9$

Only the second solution makes sense. Therefore, the projectile travels $latex 18.9\hspace{1mm}s$ from the cliff to the sea.

b) The maximum height with respect to the starting point on the cliff, can be calculated through:

$$y_{max}=\frac{v_{0y}^2}{2g}$$

Substituting values:

$$y_{max}=\frac{\left(84.5\hspace{1mm}\dfrac{m}{s}\right)^2}{2\times 9.8\hspace{1mm}\dfrac{m}{s^2}}=364.3\hspace{1mm}m $$

To find the maximum height above sea level, the 150 m height of the cliff is added:

$$H_{max}=364.3\hspace{1mm}m+150.0\hspace{1mm}m=514.3\hspace{1mm}m$$

c) The horizontal range is determined from:

$$x_{max}=v_{0x}\cdot t_{flight}=$$

$$x_{max}=181.3\hspace{1mm}\frac{m}{s}\times18.9\hspace{1mm}s=3426.6\hspace{1mm}m$$

a) The components of the initial velocity of the ice chunk are:

$latex v_{0x}=8.6\times\cos 37º \hspace{1mm}\dfrac{m}{s}=6.9\hspace{1mm}\dfrac{m}{s}$

$latex v_{0y}=-8.6\times\sin 37º \hspace{1mm}\dfrac{m}{s}=-5.2\hspace{1mm}\dfrac{m}{s}$

The vertical component is negative since the roof is inclined downward.

The origin of coordinates is taken right at the edge of the roof, where the piece of ice starts its parabolic motion. The vertical downward direction is negative.

If the person measures 1.80 m, measured from the ground up, it means that the top of his head is in the $latex position ~y=(1.8-10)\hspace{1mm}m = -8.2\hspace{1mm}m$ , with respect to the origin of coordinates.

In addition, the statement says that the man is at a horizontal distance from the edge of the roof equal to 1.2 m. Therefore, the coordinates of the top of his head are:

$$(1.2;-8.2)\hspace{1mm}m$$

If this point belongs to the parabolic trajectory of the ice chunk, it hits the man, but if it does not belong, it does not hit him.

The position of the ice chunk at any instant $latex t$ is given by:

$latex x(t)=6.9t$

$latex y(t)=-5.2t-4.9t^2$

Let's find out at what instant of time the piece of ice is located at the position $latexx=1.2\hspace{1mm} m$:

$$t=\frac{1.2\hspace{1mm}m}{6.9\hspace{1mm}\dfrac{m}{s}}= 0.174\hspace{1mm}s$$

If the vertical position at this instant is equal to -8.2 m, it falls on the man's head, otherwise it does not hit him:

$latex y(0.174)=[-5.2\times 0.174 -4.9(0.174)^2]\hspace{1mm}=-1.05\hspace{1mm} m$

It is concluded that it does not hit him. In fact, the piece of ice is quite a distance above the man's head and still needs to reach the ground, in the position $latex y = -10\hspace{1mm}m$.

b) To know at what point on the ground the piece of ice lands, it is necessary to know the time of flight, putting $latex y =-10\hspace{1mm}m$ in the equation for the vertical position:

$latex -10=-5.2t-4.9t^2$

A second-degree equation for time is obtained:

$$ 4.9t^2+5.2t+10=0$$

The positive solution is:

$$t=0.99332\hspace{1mm}s$$

For this time, the horizontal coordinate of the ice chunk is:

$$ x(t)=6.9\times 0.99332\hspace{1mm}m=6.9\hspace{1mm}m$$

The person had no reason to worry about the ice falling on him, as it actually landed much farther ahead.

a) The origin of the coordinate system is located at the launch point. At the unknown instant for which the velocity is given, the projectile must have reached its maximum height with respect to the origin, since the vertical component of the velocity is zero.

Given that:

$$v_y^2=v_{0y}^2-2gy_{max}\Rightarrow y_{max}=\frac{v_{0y}^2}{2g}$$

From the initial velocity vector, which we know is $latex \vec {v}_0=10\hat i +5\hat j\hspace{1mm}\dfrac{m}{s}$, it follows that:

$latex v_{0x}=10 \hspace{1mm}\dfrac{m}{s}$

$latex v_{0y}=5 \hspace{1mm}\dfrac{m}{s}$

Therefore, the vertical position at that time is:

$$y=y_{max}=\frac{(5 \hspace{1mm}\dfrac{m}{s})^2}{2\times 9.8 \hspace{1mm}\dfrac{m}{s^2}}=1.3\hspace{1mm}m$$

For the horizontal position:

$$x=v_{0x}\cdot t$$

The time is the maximum time, which is given by:

$$t_{max}=\frac{v_{0y}}{g}=\frac{5 \hspace{1mm}\dfrac{m}{s}}{9.8 \hspace{1mm}\dfrac{m}{s^2}}=0.51 \hspace{1mm}s$$

Which leads to:

$$x=10 \hspace{1mm}\dfrac{m}{s}\times 0.51 \hspace{1mm}s=5.1 \hspace{1mm}m$$

Putting together the previous results in the position vector $latex \vec r(t)$, we have:

$$\vec r(0.51)= \left(5.1 \hspace{1mm}\hat i +1.3\hspace{1mm}\hat j\right)\hspace{1mm}m$$

This is the position of the object with respect to the origin in t=0.51 s.

b) The launch angle $latex \theta$ is obtained from:

$$\theta =\arctan\left(\frac{v_{0y}}{v_{0x}}\right)=\arctan\left(\frac{5}{10}\right)=26.6º$$

c) The velocity vector $latex \vec v$ is given by:

$$\vec v(t) = v_x\hat i +v_y\hat j=v_{0x}\hat i+(v_{0y}-gt)\hat j$$

After t = 1.5 s, the vertical component of the velocity is:

$$v_y(1.5)=5-(9.8\times 1.5)\hspace{1mm}\frac{m}{s}=-9.7\hspace{1mm}\frac{m}{s}$$

Therefore:

$$\vec v(1.5) = (10.0\hat i -9.7\hat j)\frac{m}{s}$$

The object is going down, since the t= 1.5 s is greater than the maximum time. Furthermore, it can be seen that the sign of the vertical component of the velocity is negative.

d) To calculate the horizontal range, we need the flight time, which is not equal to twice the maximum time, since the launch is not at level, since the statement warns that it was projected at a height of 6 m above the ground.

In this case, the flight time is calculated when the mobile is at the vertical coordinate $latex y = -6\hspace{1mm}m$, which is assigned a negative sign because it is below the launch point

$$y(t)=v_{0y}t-\frac{1}{2}gt^2$$

Substituting values:

$$-6=5t-4.9t^2\Rightarrow 4.9t^2-5t-6=0$$

The positive solution is:

$$t=1.7\hspace{1mm}s$$

The maximum range is given by:

$$x_{max}=v_{0x}\cdot t_{flight}=10\hspace{1mm}\frac{m}{s}\times1.7\hspace{1mm}s= 17\hspace{1mm}m$$

a) The launch point is located at the origin of the coordinate system, so it is at the point $latex (0.0)$. The ball passes just over the edge of the wall.

The coordinates of this point, with respect to the launching point, are: $latex (10.18-2)=(10.16)\hspace{1mm}m$ (2 m must be subtracted from the height of the wall, since the launching point is 2 m above the ground.

We start from the position equations:

$latex x(t)=v_{0x}t$

$latex y(t)=v_{0y}t-\frac{1}{2}gt^2$

Given that the point $latex (10,16)\hspace{1mm}m$ and the horizontal component of the velocity equal to 15 m/s are known, we obtain:

$latex 10=15t$

$latex 16=v_{0y}t-4.9t^2$

From the first equation results:

$$t=\frac{10}{15}\hspace{1mm}s=0.67\hspace{1mm}s$$

This is the time it takes for the ball to reach the edge of the wall.

Substituting it in the second equation, the vertical component of the velocity is obtained:

$latex v_{0y}=\dfrac{16+4.9t^2}{t}=\dfrac{16+4.9\times 0.67^2}{0.67}\hspace{1mm}\dfrac{m}{s}=27.2\hspace{1mm}\dfrac{m}{s}$

b) The launch angle of the ball is:

$$\arctan\left(\frac{v_{0y}}{v_{0x}}\right)=\arctan\left(\frac{27.2}{15}\right)=61.1º$$

c) The vertical coordinate of the landing point is $latex y=-2\hspace{1mm}m$, therefore, the flight time is calculated by solving the following quadratic equation:

$$-2=27.2t-4.9t^2$$

$$4.9t^2-27.2t-2=0$$

The positive solution is:

$$t_{flight}=5.6\hspace{1mm}s$$

d) The horizontal range is given by:

$latex x_{max}=v_{0x}\cdot t_{flight}$

Therefore:

$$x_{max}=15\hspace{1mm}\dfrac{m}{s}\times 5.6\hspace{1mm}s=84\hspace{1mm}m$$

Since the wall is 10 m from the launching point, measured horizontally, the ball falls at the next horizontal distance from it:

$$x_{wall}=(84 -10)\hspace{1mm}m=74\hspace{1mm}m$$

e) The horizontal component of the velocity remains unchanged during the whole movement, then:

$$v_x=v_{0x}=15\hspace{1mm}\frac{m}{s}$$

The vertical component is calculated knowing the flight time:

$$v_y=v_{0y}-9.8t=(27.2-9.8\times 5.6)\hspace{1mm}\frac{m}{s}=-27.7\hspace{1mm}\frac{m}{s}$$

therefore:

$$\vec v_{final}=\left(15\hspace{1mm}\hat i-27.7\hspace{1mm}\hat j\right)\hspace{1mm}\frac{m}{s}$$

a) Since the launch is level, the parabolic trajectory is symmetrical and $latex t_{flight}=2t_{max}$.

The maximum time is calculated by:

$$t_{max}=\frac{v_{0y}}{g}$$

Substituting into the equation for the horizontal range:

$$x_{max}=v_{0x}\left(2\frac{v_{0y}}{g}\right)=2v_{0}\cos\theta\frac{v_{0y}}{g}=\frac{2v_{0}^2\cos\theta\sin\theta}{g}=\frac{v_{0}^2\sin2\theta}{g}$$

The maximum height is:

$$y_{max}=\frac{v_{0y}^2}{2g}=\frac{v_{0}^2(\sin\theta)^2}{2g}$$

According to the statement, the horizontal range is 5 times the maximum height, so that we obtain:

$$x_{max}=\frac{2v_{0}^2\cos\theta\sin\theta}{g}=\frac{5v_{0}^2(\sin\theta)^2}{2g}$$.

Simplifying:

$$2\cos\theta=\frac{5}{2}\sin\theta\Rightarrow 5\sin\theta=4\cos\theta\Rightarrow \tan\theta=\frac{4}{5}$$

$$\theta=38.7º$$

b) Starting from:

$$x_{max}=\frac{2v_{0}^2\cos\theta\sin\theta}{g}$$

We solve for $latex v_0$ and substitute the numerical values:

$$v_{0}=\sqrt{\frac{gx_{max}}{2\cos\theta\sin\theta}}=\left(\sqrt{\frac{9.8\times 200}{2\cos 38.7º\sin 38.7º}}\right)\hspace{1mm}\frac{m}{s}=45.7\hspace{1mm}\frac{m}{s}$$

c) The flight time is calculated by substituting values in:

$$t_{flight}=2t_{max}=2\frac{v_{0y}}{g}=2\frac{v_{0y}}{g}=\frac{2\times45.7\times\sin 38.7º}{9.8}\hspace{1mm}s=5.8\hspace{1mm}s$$

a) To find out if the stones collide, first the position equations for each stone are posed.

The origin of coordinates must be placed at a convenient point. In this case, the q stone throwing point was chosen.

(Note: it can also be placed at the launching point of p, adjusting, the result will be the same).

Stone p: horizontal throw

$latex x_p=16t_p$

$latex y_p=55-4.9t_p^2$

Stone q: Throw at an angle

The components of its initial velocity are:

$latex v_{0x}=32\times\cos60º\hspace{1mm}\dfrac{m}{s}=16.0\hspace{1mm}\dfrac{m}{s}$

$latex v_{0y}=32\times\sin60º\hspace{1mm}\dfrac{m}{s}=27.7\hspace{1mm}\dfrac{m}{s}$

And their equations of position have the form:

$latex x_q=v_{0x}t_q=16t_q$

$latex y_q=v_{0y}t_q-4.9t_q^2=27.7t_q-4.9t_q^2$

If the stones meet, they will meet at the same time, so that $latex t_p=t_q=t$. Note that the horizontal components of the position of each stone are the same, since $latex x_p=x_q=16t$.

Then, the vertical components are equal:

$$y_p=y_q\Rightarrow 55-4.9t^2=27.7t-4.9t^2$$

Meaning:

$$27.7t=55\Rightarrow t=\frac{55}{27.7}\hspace{1mm}{s}=1.98 \hspace{1mm}{s} $$

The stones will collide after 1.98 sec.

b) The coordinates of the point where the stones collide are:

$$x=16\hspace{1mm}\frac{m}{s}\times 1.98 \hspace{1mm}{s}=31.75 \hspace{1mm}{m}$$

$$y=(55-4.9\times 1.98^2)\hspace{1mm}m=35.79\hspace{1mm}{m}$$

The stones are located at the point: (31.75; 35.79) m.

c) For this we, need to plot the trajectory $latex y(x)$ for each stone, in the same Cartesian system:

Trajectory of p

Stone p: horizontal throw

$latex x_p=16t\Rightarrow t=\dfrac{x}{16}$

$latex y_p=55-4.9t^2\Rightarrow y=55-4.9\left(\dfrac{x}{16}\right)^2=55-0.019x^2$

Stone q: Throw at an angle

$latex x_q=16t\Rightarrow t=\dfrac{x}{16}$

$latex y_q=\dfrac{27.7x}{16}-4.9\left(\dfrac{x}{16}\right)^2=1.73x-0.019x^2$

The graph below shows the intersection of the parabolic trajectories at the indicated point.