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 ?
Jayant Kumar
12 Years agoGrade 11
2 Answers
Arun
6 Years ago
Ball performs parabolic motion, For vertical motion, s=ut−21gt2 ∴0=10sin60ot−21×10×t2 ∴t=3s For horizontal motion, s=ut+21at2 ∴1.15=10cos60o×3−21a(3) ∴a=5m/s2
Vikas TU
6 Years 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
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