Question

1. a person dives from a board above a pool. the height of the person is given by the function h(t)=-4.9t² + 19.6, where t is the time in seconds since the stone was dropped, and 19.6 is the initial height of the person in feet above the water.

part a
how long does it take for the person to reach the surface of the pool?
part b
if the initial height of the person is decreased by 14.7 meters, how much less time in seconds will it take the person to reach the surface of the pool?

Explanation:

Step1: Set height to 0 for Part A

When the person reaches the surface of the pool, $h(t)=0$. So we set $- 4.9t^{2}+19.6 = 0$.

Step2: Solve the quadratic equation for Part A

First, rewrite the equation as $4.9t^{2}=19.6$. Then $t^{2}=\frac{19.6}{4.9}=4$. Taking the square - root of both sides, since $t\geq0$ (time cannot be negative), $t = 2$ seconds.

Step3: Set up the new equation for Part B

The new height function is $h(t)=-4.9t^{2}+14.7$. Set $h(t) = 0$, so $-4.9t^{2}+14.7=0$. Rewrite it as $4.9t^{2}=14.7$, then $t^{2}=\frac{14.7}{4.9}=3$. So $t=\sqrt{3}\approx1.73$ seconds.

Step4: Calculate the difference in time for Part B

The difference in time $\Delta t=2 - \sqrt{3}\approx2 - 1.73 = 0.27$ seconds.

Answer:

Part A: 2 seconds
Part B: Approximately 0.27 seconds