Question

A man is sitting in an open car which is travelling along a road at a speed of $30 m s^{- 1}$ . The man stands up and throws a ball at a speed of $15 m s^{- 1}$ relative to himself at an angle $θ$ to the horizontal. The moment the man throws the ball, the driver of the car begins to accelerate at a constant rate of $10 \sqrt{3} m s^{- 2}$ , in the direction in which it was initially travelling.
(a) Find the value of $θ$ if the man in the car catches the ball.
(b) Find the time at which the man catches the ball if the instant at which the ball was thrown is $t = 0$ .

Assume that the time at which the ball travels contains the line of the motion of the car and that $g = 10 m s^{- 2}$ .

Accepted Answer

You can attempt the problem in a reference frame fixed to the car or in the reference frame fixed to the ground. Fundamentally, these two reference frames are equivalent in all respects, but in the reference frame fixed to the car, the calculations are slightly easier.

(a) Analysis of the motion of the ball in the reference frame fixed to the car: Let $x$ -axis of the coordinate system be along the motion of the car and $y$ -axis along the vertical direction. Acceleration of the ball in the reference frame,

${\vec{a}}_{BR} =$ Acceleration of the ball with respect to ground $-$ Acceleration of the reference frame with respect to ground

${\vec{a}}_{BR} = (- g \hat{j}) - (a \hat{i}) = - a \hat{i} - g \hat{j}$

$\Rightarrow {(a_{BR})}_{x} = - a and {(a_{BR})}_{y} = - g$

Let $u$ be the velocity of the ball at $t = 0$ . Then $u_{x} = u cosθ$ and $u_{v} = u sinθ$
Using the equation, $s = u t + \frac{1}{2} a t^{2}$ in both $x$ - and $y$ -directions.

Along $x$ -direction: $0 = (u cosθ) t + \frac{1}{2} (- a) t^{2} \dots (1)$

At $t = t, x = 0, y = 0$ and $0 = (u sinθ) t + \frac{1}{2} (- g) t^{2} \dots (2)$
The man catches the ball.
From these equations, either $t = 0$ which refers to the initial condition or $t = \frac{2 u sinθ}{g}$ from $(2)$ and $t = \frac{2 u cosθ}{a}$ from $(1)$

$\frac{2 u sinθ}{g} = \frac{2 u cosθ}{a}$
$tanθ = \frac{g}{a} = \frac{10}{10 \sqrt{3}} = \frac{1}{\sqrt{3}}$ .

$\Rightarrow θ = 30 °$

$(b)$ Time after releasing the ball at which man catches the ball again is given as $t = \frac{2 u sinθ}{g} = \frac{2 \times 15 \times \frac{1}{2}}{10} = 1.5 s$

B M Sharma Solutions for Chapter: Kinematics II, Exercise 33: ILLUSTRATION 5.31

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 33: ILLUSTRATION 5.31

Questions from B M Sharma Solutions for Chapter: Kinematics II, Exercise 33: ILLUSTRATION 5.31 with Hints & Solutions

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