A ball is thrown toward a cliff of height (h) with a speed of30 m/s and an angle

Question

A ball is thrown toward a cliff of height (h) with a speed of30 m/s and an angle of 60 degrees above the horizontal. Itlands on the edge of the cliff 4.0s later. a) How high is the cliff? b)What was the maximum height of the ball? c)What is the ball's impact speed? A ball is thrown toward a cliff of height (h) with a speed of30 m/s and an angle of 60 degrees above the horizontal. Itlands on the edge of the cliff 4.0s later. a) How high is the cliff? b)What was the maximum height of the ball? c)What is the ball's impact speed?

Explanation / Answer

In the x direction:
a = 0
v = 30 * cos 60 = 15 m/s
x = 30 * t * cos 60 = 30 * 4 * cos 60 = 60 meters

In the y direction:
a = -g
v = -g * t + 30 * sin 60
= -9.8 * 4 + 30 * sin 60
= -13.219 m/s
y = -(1/2) * g * t^2 + 30 * t * sin 60
= -(1/2) * 9.8 * 4^2 + 30 * 4 * sin 60
= -78.4 + 103.923
= 25.523 m

So
1.) The cliff is 25.5 meters high.

2.) The ball's impact speed is 20 m/s

vf^2 = vx^2 + vy^2
= (15)^2 + (-13.219)^2
= 399.748
vf = 19.994 m/s

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