Question

1. -/3 points if a ball is thrown vertically upward with an initial velocity of 96 ft/s, then its height after t seconds is s = 96t - 16t². (consider up to be the positive direction.) (a) what is the maximum height (in ft) reached by the ball? ft (b) what is the velocity (in ft/s) of the ball when it is 128 ft above the ground on its way up? ft/s what is the velocity (in ft/s) of the ball when it is 128 ft above the ground on its way down? ft/s

Explanation:

Step1: Find the time at maximum height

The velocity function $v(t)$ is the derivative of the position - function $s(t)$. Given $s(t)=96t - 16t^{2}$, then $v(t)=s^\prime(t)=96 - 32t$. At maximum height, the velocity $v(t)=0$. So we set $96 - 32t = 0$. Solving for $t$ gives $32t=96$, so $t = 3$ seconds.

Step2: Calculate the maximum height

Substitute $t = 3$ into the position - function $s(t)$. $s(3)=96\times3-16\times3^{2}=288 - 144=144$ feet.

Step3: Find the time when the ball is at 128 ft

Set $s(t)=128$, so $96t-16t^{2}=128$. Rearrange to get $16t^{2}-96t + 128 = 0$. Divide through by 16: $t^{2}-6t + 8 = 0$. Factor the quadratic equation: $(t - 2)(t - 4)=0$. So $t = 2$ seconds (on the way up) and $t = 4$ seconds (on the way down).

Step4: Find the velocity at $t = 2$ and $t = 4$

The velocity function is $v(t)=96 - 32t$. When $t = 2$, $v(2)=96-32\times2=96 - 64 = 32$ ft/s. When $t = 4$, $v(4)=96-32\times4=96 - 128=-32$ ft/s.

Answer:

(a) 144
(b) 32
(c) - 32