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3-2.Motion in Plane
hard
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
A$5$
B$6$
C$7$
D$8$
(IIT-2011)
Solution
Ball performs parabolic motion,
For vertical motion,
$s = ut -\frac{1}{2} g t ^2$
$\therefore o =10 \sin 60^{\circ} t -\frac{1}{2} \times 10 \times t ^2$
$\therefore t =\sqrt{3} s$
For horizontal motion,
$s=u t+\frac{1}{2} a t^2$
$\therefore 1.15=10 \cos 60^{\circ} \times \sqrt{3}-\frac{1}{2} a (3)$
$\therefore a=5 \ m / s ^2$
For vertical motion,
$s = ut -\frac{1}{2} g t ^2$
$\therefore o =10 \sin 60^{\circ} t -\frac{1}{2} \times 10 \times t ^2$
$\therefore t =\sqrt{3} s$
For horizontal motion,
$s=u t+\frac{1}{2} a t^2$
$\therefore 1.15=10 \cos 60^{\circ} \times \sqrt{3}-\frac{1}{2} a (3)$
$\therefore a=5 \ m / s ^2$
Standard 11
Physics
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