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 to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball

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 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 ____. [2011]

Smart Achievers · Accepted Answer

(5)

$Here, u = 10 m/s, θ = 60 °$

$∴$ $u_{x} = u \cos θ = 10 \times \cos 60 ° = 5 m/s$

$and u_{y} = u \sin θ = 10 \times \sin 60 ° = 5 \sqrt{3} m/s$

$∴$ $t = \frac{2 u_{y}}{g} = \frac{2 \times 5 \sqrt{3}}{10} = \sqrt{3} s$

$Let a be the acceleration of train.$

$1.15 = u_{x} t - \frac{1}{2} a t^{2}$

$\Rightarrow 1.15 = 5 \times \sqrt{3} - \frac{1}{2} \times a \times {(\sqrt{3})}^{2}$

$\Rightarrow \frac{3 a}{2} = 5 \sqrt{3} - 1.15 = 8.65 - 1.15 = 7.5$

$∴$ $a = 7.5 \times \frac{2}{3} = 5 {m/s}^{2}$

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Solution

Ans. (5) Here, u=10 m/s, θ=60° ∴ ux=ucosθ=10×cos60°=5 m/s and uy=usinθ=10×sin60°=53 m/s ∴ t=2uyg=2×5310=3 s Let a be the acceleration of train. 1.15=uxt-12at2 ⇒1.15=5×3-12×a×(3)2 ⇒3a2=53-1.15=8.65-1.15=7.5 ∴ a=7.5×23=5 m/s2

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