First slide

Projection Under uniform Acceleration

Question

A particle is projected horizontally from the top of a tower with velocity 20 m/s. Then the time after which its speed will be $20 \sqrt{2} m / s$ . (Take g = 10 $m / s^{2}$ )

Moderate

A

$\sqrt{2} \sec$
B

1 sec
C

2 sec
D

3 sec

Solution

After time t, horizontal component of velocity of the particle is, $V_{H}$ = 20 m/s and its vertical component of velocity is
$\begin{array}{l} V_{v} = 0 + gt = 10 t m / s . \\ ∴ V_{H}^{2} + V_{v}^{2} = {(20 \sqrt{2})}^{2} \\ \Rightarrow {(20)}^{2} + {(10 t)}^{2} = 800 \Rightarrow t = 2 \sec . \end{array}$

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