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Question

A baseball is thrown horizontally at 25 meters per second from a cliff 45 meters above level ground.
Approximately how far from base of cliff does the ball hit the ground

230m
140m
45m
75m

Answer 1

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The initial velocity is25 m/s in horizontal direction.

The component of initial velocity in the vertical direction is zero

The vertical displacement = 45 meter

Using, S=ut+(1/2)at^2

S=45 m

u=zero

a=9.8 m/s^2

The time taken to fall 45 meter is given by:

t=sq. rt.2S/a

t =sq rt 90/9.8=3.03 second

Horizontal motion is uniform with uniform velocity 25 m/s

Horizontal displacement = horizontal velocity x time

Horizontal displacement =25 x3.03=75.75 m

Horizontal displacement is nearly 75 meter.
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Answer 2

Figure out how long it takes for the ball to fall 45 meters.
s = at²/2 => t = √(2s/a) = √(90/9.8) = 3.03 s
So, traveling at 25 m/s horizontally, it travels
3.03 * 25 = 75.75 m from the base of the cliff before it hits the ground.

HTH

Doug

Answer 3

75