Question

A parachutist jumps out of an airplane and accelerates with gravity to a maximum velocity of 58.8 m/s in 6.00 seconds. She then pulls the parachute cord and after a 4.00-second constant deceleration, descends at 10.0 m/s for 60.0 seconds, reaching the ground. From what height did the parachutist jump?

Answer 1

Answer:

914m

Explanation:

We must divide the parachutist motion into three parts:

First we have free fall motion:

$y_1=\frac{v_f^2-v_0^2}{2g}\\y_2=\frac{(58.8\frac{m}{s})^2-0^2}{2(9.8\frac{m}{s^2})}=176.4m$

Then, we have a uniformly accelerated motion. The initial speed is this part will be the same final speed as the previous part.

$a=\frac{v_f-v_0}{t}\\a=\frac{10.0\frac{m}{s}-58.8\frac{m}{s}}{4s}=-12.2\frac{m}{s^2}\\y_2=\frac{v_f^2-v_0^2}{2a}\\y_2=\frac{(10.0\frac{m}{s})^2-(58.8\frac{m}{s})^2}{2(-12.2\frac{m}{s^2})}=137.6m$

Finally, we have a uniform linear motion:

$y_3=v*t\\y_3=10\frac{m}{s}*60s=600m$

The total heigh will be the sum of all heights:

$y=y_1+y_2+y_3\\y=176.4m+137.6m+600m=914m$