Answer to Question #157930 in Mechanics | Relativity for John Bittinger

Question

A skydiver steps out of a perfectly good airplane. How far did the skydiver

fall in 5 sec? What was the skydiver’s velocity after 5 sec? Create

position-time, velocity-time, and acceleration-time curves for the first five

sec of the fall. (Assume no other influences are acting on the skydiver,

acceleration due to gravity is –10 m/s2, and start at time 0 with 0.25-sec

increments for the plots.)

Answer 1

Explanations & Calculations

• It has to be assumed that the skydiver has no horizontal velocity just after the jump to solve this problem as no evidence for such a velocity hence his initial speed is zero.
• And just after the jump, the only force that acts on him is his body weight which is due to gravity.
• Then he gains only a vertical velocity due to the acceleration caused by his body weight.($\small F=mg=ma\to a=g$ )

• His instantaneous velocity then changes according to the equation, $\small V=U+at\to V=gt$
• Then his position/displacement changes according to the equation $\small S=ut+\frac{1}{2}at^2\to S=\frac{1}{2}gt^2$
• And the acceleration does not change during the free fall thus $\small a=g=10ms^{-2}$: considered positive downwards.

• And it is easy to plot these using excel as the 0.25s intervals are to be considered.