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

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 . 

Answer 2

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