Question

A
person is standing on top of a 57-m cliff. He throws a ball
straight up right at the very edge of the cliff so that when it
falls it lands at the base of the cliff. The speed of the ball as
it leaves his hand is 22.5 m/s.

A. What is the ball's maximum height?

B. How long is the ball in the air?

C. At what heights (there are two) is the speed of the ball
10.0 m/s?

D. What is the speed of the ball when it hits the
ground?

Answer #1

A)

We know formula

v² = u² + 2as

At maximum height v = 0

s is height above cliff

Acceleratiin due to gravity is opposite to initial velocity 'u' -ve

u² = 2 g s

s = 25.83 m

Maximum height = 57+s = 82.83 m

B)

Now

Total displacement s

s = - 57

Accelration a = - g

u = 22.5 m/s

total time in air t

s = ut + at²/2

-57 = 22.5t - 4.9t²

Solving quadratic equation

t = 6.41 s

C)

v² = u² + 2as

10² = 22.5² - 2 x 9.8 x s

s = 20.73 m

D)

v² = 22.5² - 2 x 9.8 x (-57)

= 40.3 m/s

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