Question

a ball is thrown upward from a height of 152 feet above the ground, with an initial velocity of 96 feet per second. from physics it is known that the velocity at time t is v(t)=96 - 32t feet per second
a) find s(t), the function giving the height of the ball at time t.
b) how long will the ball take to reach the ground?
c) how high will the ball go?

Explanation:

Step1: Find the position function \(s(t)\)

We know that \(v(t)=s^{\prime}(t) = 96 - 32t\). Integrating \(v(t)\) with respect to \(t\):

$$
LATEXBLOCK0
$$

Since \(s(0) = 152\) (initial height), then \(C = 152\). So \(s(t)=-16t^{2}+96t + 152\)

Step2: Find the time when the ball hits the ground (for part b)

Set \(s(t)=0\), so \(-16t^{2}+96t + 152=0\). Divide through by \(-4\) to get \(4t^{2}-24t - 38 = 0\). Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a = 4\), \(b=-24\), \(c=-38\)

$$
LATEXBLOCK1
$$

We take the positive root \(t=3+\frac{\sqrt{74}}{2}\approx3 + 4.30 = 7.30\) seconds

Step3: Find the maximum height (for part c)

The velocity at the maximum - height is \(v(t)=0\). Set \(96-32t = 0\), then \(t = 3\) seconds.
Substitute \(t = 3\) into \(s(t)\): \(s(3)=-16\times3^{2}+96\times3 + 152=-144 + 288+152=296\) feet

Answer:

a) \(s(t)=-16t^{2}+96t + 152\)
b) Approximately \(7.30\) seconds
c) \(296\) feet