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A golf ball rolls off a horizontal cliff with an initial speed of 11.0 m/s. The...

A golf ball rolls off a horizontal cliff with an initial speed of 11.0 m/s. The ball falls a vertical distance of 13.7 m into a lake below. (a) How much time does the ball spend in the air? (b) What is the speed v of the ball just before it strikes the water? (Write on paper)

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Answer #1

Initial speed u = 11.0 m / s
vertical distance S = 13.7 m
(a). time taken to spend in air t = sqrt [ 2S / g ]

t=sqrt [2*13.7/9.8]
= 1.672 s
(b). horizontal velocity of the ball when it strikes the water U = u = 11.0 m / s
vertital velocity of the ball when it strikes the water

U ' = gt

=9.8*1.672

= 16.38 m/ s
the speed v of the ball just before it strikes the water v = sqrt [ U^ 2+ U'^ 2]

=sqrt [11^2 + 16.38^2]
v = 19.73 m / s (Answer)

I hope help you !!

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