Answer:
80 ft/s
Step-by-step explanation:
Given that,
A ball is thrown vertically upward from the top of a 96-ft tower, with an initial velocity of 16 ft/sec. Its position function is given by :
[tex]s(t)=-16t^2 + 16t + 96[/tex] ...(1)
When it hits the ground, s(t) = 0
[tex]-16t^2 + 16t + 96=0\\\\t=3\ s[/tex]
At t = 3 s it will hit the fround.
We need to find its velocity when it hits the ground.
Differentiate equation (1) to find its velocity.
[tex]v=\dfrac{ds(t)}{dt}\\\\v=\dfrac{d(-16t^2 + 16t + 96)}{dt}\\\\v=-32t+16[/tex]...(2)
Put t = 3 in equation (2)
[tex]v=-32(3)+16\\\\v=-80\ ft/s[/tex]
Hence, when it will ground its velocity is 80 ft/s.
