Question

a diver jumps from a platform. the height above water in feet after t seconds is given by h(t) = -16t² + vt + 32. if the diver hits the water after 3 seconds, what was the initial velocity v?

Explanation:

Step1: Find the derivative of the height function

The height function is \( h(t)=-16t^{2}+vt + 32 \). The derivative \( h^\prime(t)\) (which gives the velocity function) using the power rule \((x^n)^\prime=nx^{n - 1}\) is \( h^\prime(t)=-32t + v\).

Step2: Determine the time when the diver hits the water

When the diver hits the water, \( h(t)=0\). We know \( t = 3\) seconds (since we are asked about the velocity when the diver hits the water after 3 seconds).

Step3: Calculate the velocity at \(t = 3\)

Substitute \(t = 3\) into the velocity function \(v(t)=h^\prime(t)=-32t + v\). But we can also use the fact that when \(t = 3\), \(h(3)=0\). First, \(h(3)=-16\times3^{2}+v\times3 + 32=0\).

$$
LATEXBLOCK0
$$

Another way: Using the velocity - time formula. The general formula for the velocity of a free - falling object (with initial velocity \(v\)) is \(v(t)=v-32t\). When \(t = 3\), if we use the fact that the height \(h(t)=-16t^{2}+vt + 32\) and \(h(3) = 0\) (as above) or directly from the derivative \(v(t)=-32t + v\). If we assume the standard kinematic approach (derived from calculus, where \(a=-32\) ft/s², \(v(t)=v_0+at\), here \(a=-32\), \(t = 3\)).

Answer:

\(v = 37.3\)