a skydiver jumps from an airplane at an altitude of 2,500 ft. he falls under the force of gravity until he…

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

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

Answer

Explanation:

Step1: Substitute values into the formula

Given (h(t)=- 16t^{2}+vt + h_{0}), (v = 0) (initial velocity), (h_{0}=2500) (initial height), and (h(t)=1000) (height when parachute opens). Substitute into the formula: (1000=-16t^{2}+0\times t + 2500).

Step2: Simplify the equation

(1000=-16t^{2}+2500). Move terms: (16t^{2}=2500 - 1000). (16t^{2}=1500).

Step3: Solve for (t)

(t^{2}=\frac{1500}{16}). (t=\sqrt{\frac{1500}{16}}). (t=\frac{\sqrt{1500}}{4}). Since (\sqrt{1500}=\sqrt{100\times15} = 10\sqrt{15}\approx10\times3.87 = 38.7). (t=\frac{38.7}{4}\approx9.7).

Answer:

(9.7s)