Answer:
914m
Explanation:
We must divide the parachutist motion into three parts:
First we have free fall motion:
[tex]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[/tex]
Then, we have a uniformly accelerated motion. The initial speed is this part will be the same final speed as the previous part.
[tex]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[/tex]
Finally, we have a uniform linear motion:
[tex]y_3=v*t\\y_3=10\frac{m}{s}*60s=600m[/tex]
The total heigh will be the sum of all heights:
[tex]y=y_1+y_2+y_3\\y=176.4m+137.6m+600m=914m[/tex]
