Question

A truck is moving horizontally with velocity \[v\]. A ball is thrown vertically upward with a velocity \[u\] at an angle \[\theta \] to the horizontal from the truck relative to it. The horizontal range of the ball relative to the ground is
A. \[\dfrac{{2 \times \left( {v + u\cos \theta } \right)u\sin \theta }}{g}\]
B. \[\dfrac{{2 \times \left( {v + u\sin \theta } \right)u\cos \theta }}{g}\]
C. \[\dfrac{{2 \times \left( {v - u\cos \theta } \right)u\sin \theta }}{g}\]
D. \[\dfrac{{2 \times \left( {v - u\sin \theta } \right)u\cos \theta }}{g}\]

Answer 1

Hint:Use the first kinematic equation. Also use the formula for velocity of an object. Using a kinematic equation for an object in free fall, calculate the time required to reach the maximum height for the ball. Then use the formula for velocity and calculate the horizontal range of the ball relative to the ground taking the velocity of the ball as the sum of the velocity of the truck and horizontal component of velocity of the ball.

Formulae used:
The first kinematic equation is
\[v = u + at\] …… (1)
Here, \[v\] is the final velocity of an object, \[u\] is initial velocity of the object, \[a\] is acceleration of the object and \[t\] is time.
The velocity \[v\] of an object is
\[v = \dfrac{s}{t}\] …… (2)
Here, \[s\] is the displacement of the object and \[t\] is time.

Complete step by step answer:
We have given that the velocity of the truck moving horizontally is \[v\].A ball is thrown upward with a velocity \[u\] making an angle \[\theta \] with the horizontal.We have asked to calculate the horizontal range of the ball relative to the ground.Let us first calculate the time required for the motion of the ball at a point where the velocity of the ball is zero or the time required to reach the ball to maximum height with the given velocity.Rewrite equation (1) for the upward motion of the ball.
\[{u_{yf}} = {u_{yi}} - g\dfrac{T}{2}\]
Here, \[{u_{yf}}\] is the vertical component of final velocity of the ball when it reaches maximum height, \[{u_{yi}}\] is the vertical component of initial velocity of the ball and \[\dfrac{T}{2}\] is the time required for the ball to the maximum height which is half of the time of flight of the ball.

The final velocity of the ball when it reaches the maximum height is zero.
\[{u_{yf}} = 0\,{\text{m/s}}\]
Substitute \[0\,{\text{m/s}}\] for \[{u_{yf}}\] and \[u\sin \theta \] for \[{u_{yi}}\] in the above kinematic equation.
\[\left( {0\,{\text{m/s}}} \right) = u\sin \theta - g\dfrac{T}{2}\]
\[ \Rightarrow u\sin \theta = g\dfrac{T}{2}\]
\[ \Rightarrow T = \dfrac{{2u\sin \theta }}{g}\]
Hence, the time required for the ball to reach the maximum height is \[\dfrac{{2u\sin \theta }}{g}\].

Now we can calculate the horizontal range of the ball using equation (2).
Substitute \[{u_x} + v\] for \[v\], \[R\] for \[s\] and \[T\] for \[t\] in equation (2).
\[{u_x} + v = \dfrac{R}{T}\]
\[ \Rightarrow R = \left( {{u_x} + v} \right)T\]
\[ \Rightarrow R = \left( {u\cos \theta + v} \right)T\]
Substitute \[\dfrac{{2u\sin \theta }}{g}\] for \[T\] in the above equation.
\[ \Rightarrow R = \left( {u\cos \theta + v} \right)\left( {\dfrac{{2u\sin \theta }}{g}} \right)\]
\[ \Rightarrow R = \dfrac{{2 \times \left( {u\cos \theta + v} \right)u\sin \theta }}{g}\]
Therefore, the horizontal range of the ball relative to ground is \[\dfrac{{2 \times \left( {u\cos \theta + v} \right)u\sin \theta }}{g}\].

Hence, the correct option is A.

Note: The students should keep in mind that we have considered the motion of the ball as projectile motion. Hence, one can use the time required to reach the ball to the maximum height as the half of the time of flight of the ball in projectile motion. Also the students should not forget to use the horizontal component of the velocity of the ball in the formula for velocity as the range of the ball and velocity of the truck are in horizontal direction.