Question

sharon is jumping from an 18 - foot diving board with an initial upward velocity of 4 ft/s. when sharon jumps, megan throws a beach ball up to sharon with an initial upward velocity of 16 ft/s from a height 5 feet off the ground. to the nearest hundredth of a second, how long after she jumps does the ball reach sharon?
0.65 seconds
0.92 seconds
1.08 seconds
1.15 seconds

Explanation:

Step1: Write the height - time equations

The equation for the height $h$ of an object in vertical - motion under the influence of gravity is $h = h_0+v_0t-\frac{1}{2}gt^2$, where $h_0$ is the initial height, $v_0$ is the initial velocity, $t$ is the time, and $g$ is the acceleration due to gravity. The value of $g = 32$ ft/s².
For Sharon, $h_1=18 + 4t-16t^2$ (since $h_0 = 18$ ft and $v_0 = 4$ ft/s).
For the beach - ball, $h_2=5 + 16t-16t^2$ (since $h_0 = 5$ ft and $v_0 = 16$ ft/s).

Step2: Set the two height equations equal

When the ball reaches Sharon, $h_1=h_2$.
\[

$$\begin{align*} 18 + 4t-16t^2&=5 + 16t-16t^2\\ 18 + 4t-16t^2-(5 + 16t-16t^2)&=0\\ 18 + 4t-16t^2 - 5-16t + 16t^2&=0\\ (18 - 5)+(4t-16t)+(-16t^2 + 16t^2)&=0\\ 13-12t&=0 \end{align*}$$

\]

Step3: Solve for $t$

\[

$$\begin{align*} 12t&=13\\ t&=\frac{13}{12}\approx1.08 \end{align*}$$

\]

Answer:

C. 1.08 seconds