Question

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 60° to the horizontal. The boy has to move a distance of 1.15m inside the train to catch the ball when it comes back to the same height. The acceleration of the train, in ms-2, is ( g = 10ms-2 )

Answer 1

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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Answer 2

Answer:

a = 5 m/s^2

Explanation:

t = 2u sin teta / g

 t = 1 2 ×10 × $\sqrt{3}$ /2  /10

         =$\sqrt{3}$ sec

S = ut +  1/2 at^2

     1.15 = 5 ×$\sqrt{3}$ - 1/2 × a × 3

      or 1.15 = 5$\sqrt{3}$ - 3a/2

      or a = 5 m/s2