A boy in a tree throws an apple upward with a speed of 8 m/s, how...

A boy in a tree throws an apple upward with a speed of 8 m/s, how long does the apple take to reach the ground , 5m below the boy? How fast is the apple moving just before it strikes the ground?

Solutions

Expert Solution

Time taken by tha apple to reach maximum height is

t = u / g = (8 m/s ) / ( 9.8 m/s² ) =0.816 s

maximum height

H = 5 m + u² / 2g = ( 8 m/s)² /2 ( 9.8 m/s² ) + 5 m = 8.265 m

time taken to reach ground from maximum height is

t' = sqrt ( 2H / g )

= sqrt ( 2 (8.265 m ) / ( 9.8 m/s² )

= 1.298 s

total time taken by the apple to reach the ground is T = 2.114 s

velocity of the apple befor reaching the ground is

v = sqrt ( 2 g H ) = sqrt ( 2 ( 9.8 m/s² ) (8.265 m ) )

= 12.727 m/s

genius_generous answered 2 hours ago

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