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# 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 ° 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 ?

Arun
25763 Points
one year ago
Ball performs parabolic motion,
For vertical motion,
s=ut21gt2
0=10sin60ot21×10×t2
t=3s
For horizontal motion,
s=ut+21at2
1.15=10cos60o×321a(3)
a=5m/s2
Vikas TU
14149 Points
one year ago
uy=10sin 60=53√m/s
⇒     t=2uyg=2×53√10=3√s
Sx=uxt+12axt2
1.15=5×t−12a×t2
1.15=5×3√−32a
3a2=5×1.73−1.15=8.65−1.15
3a2=7.5
⇒   a=153=5m/s2