Question

a skydiver jumped out of an airplane at a height of 10,000 feet. assuming the initial velocity is zero, write a model, s(t) that represents the height of the skydiver from the ground t seconds after they jump out of the airplane. then, find the height of the skydiver 7 seconds after they step out of the airplane.
formula: s(t)=\frac{1}{2}gt^{2}+v_{0}t + s_{0}
g is the gravitational constant (-32 feet per second squared)
t is time in seconds
v_{0} is the initial velocity of the object
s_{0} is the initial height of the object from the ground
(1 point)
9,888 feet
69,216 feet
9,216 feet
9,759.9 feet

Explanation:

Step1: Identify values

Given $g=- 32$, $v_0 = 0$, $s_0=10000$. The height - time model is $s(t)=\frac{1}{2}gt^{2}+v_0t + s_0$. Substituting the values, we get $s(t)=\frac{1}{2}\times(-32)t^{2}+0\times t + 10000=-16t^{2}+10000$.

Step2: Calculate height at $t = 7$

Substitute $t = 7$ into $s(t)=-16t^{2}+10000$. Then $s(7)=-16\times7^{2}+10000$. First, calculate $7^{2}=49$. Then $16\times49 = 784$. So $s(7)=-784 + 10000=9216$.

Answer:

9216 feet