Kinematics Ques 34
- A train is moving along a straight line with a constant acceleration $a$. A boy standing in the train throws a ball forward with a speed of $10 m / s$, at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by $1.15 m$ inside the train to catch the ball back at the initial height. The acceleration of the train, in $m / s^{2}$, is.
(2011)
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Answer:
Correct Answer: 34.$5 ms^{-2}$
Solution:
Formula:
- $t=T=\frac{2 u \sin \theta}{g}=\frac{2 \times 10 \times \sin 60^{\circ}}{10}=\sqrt{3} s$
Displacement of train in time $t=\frac{1}{2} a t^{2}$
Displacement of boy with respect to train $=1.15 m$
$\therefore$ Displacement of boy with respect to ground
$$ =1.15+\frac{1}{2} a t^{2} $$
Displacement of ball with respect to ground $=\left(u \cos 60^{\circ}\right) t$ To catch the ball back at initial height,
$$ \begin{aligned} 115+\frac{1}{2} a t^2 & =\left(u \cos 60^{\circ}\right) t \\ \therefore \quad 1.15+\frac{1}{2} a(\sqrt{3})^2 & =10 \times \frac{1}{2} \times \sqrt{3} \end{aligned} $$
Solving this equation, we get
$$ a=5 \mathrm{~m} / \mathrm{s}^2 $$
BETA
