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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 tangential and normal components of acceleration of stone are same is : (neglect air resistance) $g = 10 m / s^{2}$ .

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

$\frac{3}{2} \sec$

$3 \sec$

$4 \sec$

answer is C.

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Detailed Solution

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

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

$a_{t} = a_{n}$

$g \sin θ = g \cos θ \Rightarrow θ = 45^{\circ}$

$\Rightarrow V_{y} = V_{x}$

$U_{y} - g_{t} = U_{x} \Rightarrow 20 - (10) t = 10 \Rightarrow t = 1 \sec$

During downward motion $a_{t} = a_{n}$

$V_{y} = - V_{x} \Rightarrow 20 - 10 t = - 10 \Rightarrow t = 3 \sec$

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