Question

a skydiver jumps from an airplane at an altitude of 2,500 ft. he falls under the force of gravity until he opens his parachute at an altitude of 1,000 ft. using the initial velocity of zero, approximately how long does the jumper fall before he opens his chute?
for this quadratic model we will let the y - axis be the axis of symmetry.
h(t)=-16t^{2}+vt + h_{0}
2.4 s
9.7 s
12.5 s
14.8 s

Explanation:

Step1: Determine values for formula

The initial height $h_0 = 2500$ ft, the final height $h(t)=1000$ ft and the initial - velocity $v = 0$ ft/s. The height - time formula is $h(t)=-16t^{2}+vt + h_0$. Substituting the values, we get $1000=-16t^{2}+0\times t + 2500$.

Step2: Rearrange the equation

Rearrange the equation $1000=-16t^{2}+2500$ to solve for $t$. First, subtract 2500 from both sides: $1000 - 2500=-16t^{2}$, which simplifies to $- 1500=-16t^{2}$. Then divide both sides by - 16: $t^{2}=\frac{1500}{16}=93.75$.

Step3: Solve for t

Take the square - root of both sides. Since $t>0$ (time cannot be negative in this context), $t=\sqrt{93.75}\approx9.7$ s.

Answer:

9.7 s