Question

a train is moving along a straight line with a constant acceleration. a boy standing in the train throws a ball forward with a speed of 10m/s at an angle of 60degree to the horizontal. the boy has to move forward by 1.15m inside the train to catch the ball back at initial height. the acceleration of the train is????? plz giv d soln of this...

Answer 1

Let the velocity of  ball be  u₀

We know that,

$\text{time  of  flight (t)} = \frac{2u_0\sin\theta}{g} \\ \\  t = \frac{2 * 10\sin60 \textdegree}{10} \\ \\ = \frac{20 \times \frac{\sqrt{3}}{2}}{10} \\ \\ = \sqrt{3} \ sec. \\ \\ \text{Let the distance  travelled  by the ball be  x}  \\ \\ \ x =  u_0\cos\theta t + \frac{1}{2}at^{2}  \\ \\ x = 10* \cos60 \textdegree *\sqrt{3} + \frac{1}{2}a \times (\sqrt{3})^{2} \\ \\ x =  5\sqrt{3} + \frac{3a}{2}  \\ \\ \text{Since  boy  has  to move  1.5 m forward tp caught the ball} \\ \\ x = 1.15 \\ \\   5\sqrt{3} + \frac{3a}{2}  = 1.15 \\ \\ 3a = - 15.02 \\ \\ a  = - 5  \ \ m/sec$

Answer 2

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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