Question

A person is standing on a truck moving with a velocity of $14.7m{s^{ - 1}}$ on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved $58.8m.$ Find the speed and angle of projection:
(A) As seen from the truck
(B) As seen from the road

Answer 1

In order to find the speed and angle of projection of the ball thrown, calculate the time by which the ball reached down. Then calculate the velocity of the ball in the upward direction. From the vertical and horizontal speed, we can calculate the angle and speed of the projection.

Complete step by step solution:
Let’s define the data given in the question.
Velocity of the truck, ${v_t}$ = $14.7m{s^{ - 1}}$
Distance covered by the truck, $s$ = $58.8m.$
We need to find the speed and angle of projection of the ball in two different angles: as seen from the truck and as seen from the road.
We are discussing the first case: as seen from the truck
When we look at the ball from the truck, the ball goes upward from the truck and reaches the truck itself.
So it seem like the ball go vertically upwards, that is angle = $90^\circ$
Now we are calculating the time taken by the truck to cover the distance.
Time taken by the truck,
$t = \dfrac{s}{{{v_t}}}$
$\Rightarrow t = \dfrac{{58.8}}{{14.7}} = 4s$
We know, the time taken by the ball to reach down to the ground,
$t = \dfrac{{2u\sin \theta }}{g}$
$\Rightarrow u = \dfrac{{tg}}{{2\sin \theta }}$
Where, $u$ is the speed of projection
$\theta$ is the angle of projection
$g$ is the gravitational constant
Applying the known values to this equation we get,
$\Rightarrow u = \dfrac{{4 \times 9.8}}{{2\sin 90}}$
$\Rightarrow u = \dfrac{{4 \times 9.8}}{{2 \times 1}} = 2 \times 9.8 = 19.6m{s^{ - 1}}$
That is, when the ball is seen from the truck, it will have a speed of $19.6m{s^{ - 1}}$ vertically upwards.
We are discussing the second case: as seen from the road:
At this time the ball will have two speeds: the vertical speed of projection and horizontal speed.
The vertical speed of projection is $u = 19.6m{s^{ - 1}}$
The horizontal speed is given by the truck, ${v_t} = 14.7m{s^{ - 1}}$
So the speed of projection,$v = \sqrt {{{19.6}^2} + {{14.7}^2}}$
$\Rightarrow v = \sqrt {384.16 + 216.09}$
$\Rightarrow v = \sqrt {600.25}$
$\Rightarrow v = 24.5m{s^{ - 1}}$
And angle of projection,$\theta = {\tan ^{ - 1}}\left( {\dfrac{{19.46}}{{14.7}}} \right)$
$\Rightarrow \theta = {\tan ^{ - 1}}1.33 = 53^\circ$
$\theta = 53^\circ$
Angle of projection, $\theta = 53^\circ$ with horizontal.
That is, when the ball is seen from the truck, it will have a speed of $24.5m{s^{ - 1}}$ at angle $53^\circ$ with horizontal.

Note: When an object is thrown upwards from a moving vehicle or from any moving position, the object will have the same velocity and direction of the vehicle at the moment of release. This is due to the inertia of the object caused by the movement of the vehicle.