If you jump out of an airplane, prior to opening your parachute, your downward v

Question

If you jump out of an airplane, prior to opening your parachute, your downward velocity t seconds after the jump is approximate by v(t) = 49(1 - e-0.2t) given in meters per second. Assume that you jump out of an airplane at a height of 1500 meters above ground and open your parachute 25 seconds after you have jumped. How high above the ground will you be 10 seconds after you jumped? If it takes an additional 250 meters to reach a safe landing speed after you open your parachute, will you land safely on the ground? Please justify your answer using calculus.

Explanation / Answer

v(t) = 49(1 - e^(-0.2t)) = 49 - 49e^(-0.2t)

Velocity is downward.

s(t) = [int ] -v(t) dt = [int ] -(49 - 49e^(-0.2t)) dt

= -49t - 245e^(-0.2t) + C

s(0) = 1500 m

=> 0 - 245 + C = 1500

C = 1745

s(t) = -49t - 245e^(-0.2t) + 1745

a) s(10) = -49*10 - 245e^(-0.2*10) + 1745

= 1221.843 m

b) s(25) = -49*25 - 245e^(-0.2*25) + 1745

= 518.35 m

An additional 250 m is required for safe landing, and we have 518.35 metres.

=> We will land safely on the ground.

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