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A stone is projected from level ground at t = 0 sec such that its horizontal and vertical components of initial velocity are 10 m/s and 20 m/s respectively. Then the instant of time at which magnitude of tangential and magnitude of normal components of acceleration of stone are same is : (neglect air resistance) $g = 10 m / s^{2} .$

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1 sec

4 sec

3 sec

$\frac{1}{2} \sec$

answer is B, C.

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Detailed Solution

Tangential acceleration $= a_{t} = g \sin θ$

Normal acceleration $= a_{n} = g \cos θ$

$a_{t} = a_{n}$

$g \sin θ = g \cos θ \Rightarrow θ = 45 ° \Rightarrow V_{y} = V_{x}$

$u_{y} - g t = u_{x}, 20 - (10) t = 10, t = 1 s e c .$

During downward motion $a_{t} = a_{n}, V_{y} - V_{x}, 20 - 10 t = - 10 \Rightarrow t = 3 s e c .$

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