Question

suppose a baseball is thrown vertically upward from the ground with an initial velocity of v0 ft/s. its height above the ground after t seconds is given by s(t)= - 16t^2 + v0t. determine the initial velocity of the ball if it reaches a high point of 86 ft. the initial velocity is (round to one decimal place as needed.)

Explanation:

Step1: At the high - point, velocity is 0.

The velocity function $v(t)$ is the derivative of the height function $s(t)=-16t^{2}+v_0t$. Using the power rule, $v(t)=s^\prime(t)=-32t + v_0$. When the ball reaches its high - point, $v(t) = 0$. So, $t=\frac{v_0}{32}$.

Step2: Substitute $t=\frac{v_0}{32}$ into the height function.

Substitute $t = \frac{v_0}{32}$ into $s(t)=-16t^{2}+v_0t$. We get $s(\frac{v_0}{32})=-16(\frac{v_0}{32})^{2}+v_0(\frac{v_0}{32})$.

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Step3: Solve for $v_0$ given the height at the high - point.

Since the high - point $s = 86$ ft, we set $\frac{v_0^{2}}{64}=86$. Then $v_0^{2}=86\times64 = 5504$. So, $v_0=\sqrt{5504}\approx74.2$ ft/s.

Answer:

$74.2$