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. Approximately how long does the jumper fall before he opens his chute?
h(t) = -16t2 + vt + h0
A. 2.4 s
B. 9.7 s
C. 12.5 s
D. 14.8 s
To determine the time it takes for the skydiver to fall from an altitude of 2,500 ft to 1,000 ft, we can use the equation for the height of an object in freefall:
h(t) = -16t^2 + vt + h0
where:
h(t) is the height at time t
v is the initial velocity (0 since the skydiver starts from rest)
h0 is the initial height (2,500 ft)
In this case, we want to find the time it takes for the skydiver to reach a height of 1,000 ft, so we can set h(t) equal to 1,000 and solve for t.
1,000 = -16t^2 + 0t + 2,500
Rearranging the equation:
16t^2 = 2,500 - 1,000
16t^2 = 1,500
t^2 = 1,500 / 16
t^2 = 93.75
Taking the square root of both sides:
t ≈ √93.75
t ≈ 9.68
Therefore, it takes approximately 9.7 seconds for the skydiver to fall from an altitude of 2,500 ft to 1,000 ft.
So, the correct answer is B. 9.7 s.
hf=hi+vi^t-4.9t^2
1000=2500+0-4.9t^2
t= sqrt (1500/4.9)