Two boys are throwing a baseball back and forth. The ball is 4 ft above the ground when it leaves one child’s hand with an upward velocity of 36 ft/s. If acceleration due to gravity is –16 ft/s2, how high above the ground is the ball 2 s after it is thrown?

h(t)=at^2+vt+h0 <-- formula

2 answers

g = -32 ft/s^2
it is -32/2 which is -16
and
h(2) = (1/2)(-32) t^2 + 36 t + 4

h(2) = -16 (2)^2 + 36 (2) + 4

h(2) = -64 + 72 + 4

h(2) = 12 feet
25ft^2