Question

Two boys are throwing baseball back and forth the ball is 4 ft above ground when it leaves one child's hand with an upward velocity of 36 ft/s. If acceleration due to gravity is -16 ft/s squared how high above ground is the ball 2 seconds after it is thrown

Answer 1

formula : s = ut + 0.5at^2
consider only in vertical direction
h = 36(2) + 0.5(-16)(2^2) = 40 ft
as initial height is 4
so total height = 4 + 40 = 44 ft

Answer 2

If the ball is launched from 4 ft above the ground with an upward velocity
of 36 ft/s, and the acceleration due to gravity is -16 ft/s², then at any time
't' seconds after launch, its height above the ground is

H = -8 t² + 36 t + 4 .

After 2 seconds, H = -8(2)² + 36(2) + 4

= -32 + (72) + 4

= 44 feet above the ground .
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Now, if the boys were to move their game to the Earth, where
the acceleration due to gravity is 32 ft/s² instead of 16 ft/s² ,
the ball's behavior would be noticeably different. Two seconds
after the same launch on Earth, the height of the ball would be

H = -16 t² + 36 t + 4 .

= -16(2)² + 36(2) + 4

= -64 + 72 + 4

= 12 feet above the ground .