Question

If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t − 16t 2 . (a) what is the maximum height reached by the ball? (b) what is the velocity of the ball when it is 96 ft above the ground on its way up? on its way down?

Answer 1

Max height occurs when v = 0.
v(t) = ds(t)/dt
v(t) = 80 - 32t
0 = 80 - 32t
t = 5/2

s(5/2) = 80(5/2) - 16(5/2)^2
s(5/2) = 100
Answer: 100 ft

96 = 80t - 16t²
t = 3, 2
(80 ± √256) / 32 using the quadratic equation.

v(2) = 16
v(3) = -16