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: Recall the height formula

The height of an object in vertical motion is given by \( h(t) = -16t^2 + v_0t + h_0 \), where \( v_0 \) is the initial velocity and \( h_0 \) is the initial height.

For Sharon: \( h_S(t) = -16t^2 + 4t + 18 \) (initial height 18 ft, initial velocity 4 ft/s)

For the beach ball: \( h_B(t) = -16t^2 + 16t + 5 \) (initial height 5 ft, initial velocity 16 ft/s)

Step2: Set heights equal

To find when the ball reaches Sharon, set \( h_S(t) = h_B(t) \):

\( -16t^2 + 4t + 18 = -16t^2 + 16t + 5 \)

Step3: Simplify the equation

Add \( 16t^2 \) to both sides:

\( 4t + 18 = 16t + 5 \)

Subtract \( 4t \) from both sides:

\( 18 = 12t + 5 \)

Subtract 5 from both sides:

\( 13 = 12t \)

Step4: Solve for t

Divide both sides by 12:

\( t = \frac{13}{12} \approx 1.0833 \)

Rounding to the nearest hundredth, \( t \approx 1.08 \) seconds.

Answer:

1.08 seconds (corresponding to the option "1.08 seconds")