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13 April, 12:35

A ball is kicked horizontally at 10.0 m/s from a cliff 100m high. How far from the base of the cliff will the stone strike the ground?

a

1,000

b

60

c

10

d

3010.2

e

45

+1
Answers (1)
  1. 13 April, 12:41
    0
    We study the motion of the ball on the two coordinating needles Ox and Oy, originated on the ground. On Ox the ball has a uniform straight motion with the velocity vx = vo and the equation of motion is x = vo * t, and on the ball Oy has a movement uniformly accelerated with acceleration g and no initial velocity on the axis Oy v_oy = 0 (free fall), so the equation of motion is y = h-gt² / 2. The equation of the trajectory is obtained by eliminating the time from the two equations. We hold t = x / vo = > y = h-gx²/2vo². The trajectory is a parabola because y is a second degree function of x. Bila arrives on the ground when y = 0 = > t = √ (2h/g) = √ (2*100/9.8) = √20.4=4.51 s

    The ball will fall towards the rock at a distance d = vo * t = 10*4.51≈45 m

    The correct answer is e
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