A ball is projected from the ground at an angle θ with the horizontal. After 1s it is moving at an angle 45 ° with the horizontal and after 2s , it is moving horizontally. What is the velocity of projection of the ball?

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Quantrex Academy · Accepted Answer

Correct answer: C. 10 5 ms^ - 1

The ball will be vertical at the highest point so

{t}_{up} = \dfrac{usin\theta}{g} = 2 ⇒ usin\theta = 20 . . \ldots \ldots .(1)

∵ tan\alpha = \dfrac{usin\theta - gt}{ucos\theta}

⇒ 1 = \dfrac{20 - g ( 1 )}{ucos\theta} ⇒ ucos\theta = 10

…~. .(2)

Squaring and adding equation (1) & equation (2), we get

\left(usin\theta\right)2 + \left(\text{u}\text{cos}\theta\right)2 = {20}^{2} + {10}^{2}

⇒ u^{2} = 500

⇒ u = 10\sqrt{5}ms^{- 1}

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