A body is projected horizontally from the top of a cliff with a velocity of $9.8 m / s$ . How much time elapses before horizontal and vertical velocity becomes equal in magnitude? ( Take $g = 9.8 m s^{- 2}$ )

9.8

s

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Solution

The correct option is C 1 $s$
Horizontal velocity at any instant, $v_{x}$ = $u = 9.8 m / s$
Vertical velocity at any instant,
$v_{y} = 0 + g t = 9.8 t m / s$
To evaluate the time, equate horizontal and vertical velocity
9.8 = 9.8t
t = 1 $s$

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A body is projected horizontally from the top of a cliff with a velocity of 9.8 m/s. How much time elapses before horizontal and vertical velocity becomes equal in magnitude? ( Take g=9.8 ms−2)

The correct option is C 1 sHorizontal velocity at any instant, vx = u=9.8 m/s Vertical velocity at any instant, vy=0+gt=9.8t m/s To evaluate the time, equate horizontal and vertical velocity 9.8 = 9.8t t = 1 s

A body is projected horizontally from the top of a cliff with a velocity of $9.8 m / s$ . How much time elapses before horizontal and vertical velocity becomes equal in magnitude? ( Take $g = 9.8 m s^{- 2}$ )

The correct option is C 1 $s$
Horizontal velocity at any instant, $v_{x}$ = $u = 9.8 m / s$
Vertical velocity at any instant,
$v_{y} = 0 + g t = 9.8 t m / s$
To evaluate the time, equate horizontal and vertical velocity
9.8 = 9.8t
t = 1 $s$