A person is standing on a truck moving with a constant velocity of $14.7 m / s$ on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved $58.8 m$ .

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Angle of projection is $53 °$ with horizontal seen from ground

Speed of ball is $19.6 m / s$ upward as seen from truck

Speed of ball is $19.6 m / s$ upward as seen from road

Angle of projection is $30 °$ with horizontal seen from truck

answer is A, D.

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Detailed Solution

Angle of projection as seen from the truck $= 90 °$
Time taken by truck : $t = \frac{s}{v t} = \frac{58.8}{14.7} = 4 s$
$u = \frac{t g}{2 \sin θ} = \frac{4 \times 9.8}{2 \times 1} = 19.6 m / s$
As seen from the road $v = \sqrt{u_{x}^{2} + u_{y}^{2}} = \sqrt{v_{t}^{2} + u^{2}}$
$θ = \tan^{- 1} (\frac{19.6}{14.7}) = \tan^{- 1} (1.33) = 53 °$