Question

A ball is launched with an initial velocity of 28.4 m/s at a 45° angle from the top of a cliff that is 12.0 m above the water below. Use g = 9.80 m/s2 to simplify the calculations. Note: You could answer these questions using projectile motion methods, but try using an energy conservation approach instead. (a) What is the ball's speed when it hits the water? . m/s (b) What is the ball's speed when it reaches its maximum height? m/s (c) What is the maximum height (measured from the water) reached by the ball in its flight?

Answer 1

a)using conservation of energy

v^2 = u^2 + 2 g h

v^2 = 28.4^2 + 2* 9.8* 12

v = 32.276 m/s

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b)

only horizontal velocity left at max height

Vx = 28.4 cos 45 = 20.08 m/s

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c)

using energy conservation

0.5 m u^2 + mgh = m g H

0.5* ( 28.4 sin 45) ^2 + 9.8* 12 = 9.8*H

H = 32.576 m

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