Question

a skydiver jumps from a plane that is 5000 feet above the ground.
the skydiver opens his parachute when he is 2000 feet above the ground. substitute h=2000
in the equation
( h = -16t^2 + 5000 )
for how many seconds does the skydiver fall before opening his parachute?
______
○ about 23 seconds
○ about 14 seconds
○ about 31 seconds
○ about 8 seconds

Explanation:

Step1: Substitute h=2000 into the equation

We have the equation \( h = -16t^2 + 5000 \), and we substitute \( h = 2000 \) into it. So we get \( 2000=-16t^{2}+5000 \).

Step2: Solve for \( t^2 \)

First, we subtract 5000 from both sides of the equation: \( 2000 - 5000=-16t^{2}+5000 - 5000 \), which simplifies to \( - 3000=-16t^{2} \). Then we divide both sides by - 16: \( t^{2}=\frac{-3000}{-16}=\frac{3000}{16} = 187.5 \).

Step3: Solve for \( t \)

We take the square root of both sides. Since time \( t>0 \), we consider the positive square root. So \( t=\sqrt{187.5}\approx13.69\approx14 \) (we round to a reasonable decimal place).

Answer:

about 14 seconds