Question

a skydiver jumped out of an airplane at the height of 10,000 feet. assuming the initial velocity is zero, find the height of the skydiver 7 seconds after they step out of the airplane. write a model, h(t) that represents the height of the skydiver from the ground t seconds after they jump out of the airplane. (1 point)

9,759.9 feet

9,216 feet

69,216 feet

9,888 feet

Explanation:

Step1: Recall free - fall formula

The distance $d$ an object falls in free - fall with initial velocity $v_0 = 0$ is given by $d=\frac{1}{2}gt^{2}$, where $g = 32\ ft/s^{2}$ (acceleration due to gravity on Earth).

Step2: Calculate distance fallen in 7 seconds

Substitute $t = 7$ and $g=32$ into the formula $d=\frac{1}{2}gt^{2}$. So $d=\frac{1}{2}\times32\times7^{2}=16\times49 = 784$ feet.

Step3: Find height above ground

The initial height $h_0=10000$ feet. The height $h(t)$ above the ground $t$ seconds after jumping is $h(t)=h_0-\frac{1}{2}gt^{2}$. When $t = 7$, $h(7)=10000 - 784=9216$ feet.

Answer:

9,216 feet