sharon is jumping from an 18 - foot diving board with an initial upward velocity of 4 ft/s. when sharon…

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?\n0.65 seconds\n0.92 seconds\n1.08 seconds\n1.15 seconds

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?\n0.65 seconds\n0.92 seconds\n1.08 seconds\n1.15 seconds

Answer

Explanation:

Step1: Write height - time equations

The height - time equation for an object in vertical motion 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 $g = 32$ ft/s². For Sharon: $h_S=18 + 4t-\frac{1}{2}\times32t^2=18 + 4t-16t^2$. For the beach - ball: $h_B=5 + 16t-\frac{1}{2}\times32t^2=5 + 16t-16t^2$.

Step2: Set the two height equations equal

When the ball reaches Sharon, $h_S=h_B$. $18 + 4t-16t^2=5 + 16t-16t^2$.

Step3: Simplify the equation

Subtract $-16t^2$ from both sides of the equation: $18 + 4t=5 + 16t$.

Step4: Solve for $t$

Move the terms with $t$ to one side: $16t-4t=18 - 5$. $12t=13$. $t=\frac{13}{12}\approx1.08$ seconds.

Answer:

1.08 seconds