Question

A 70 kg skydiver falls from an altitude of 10 km to 1 km before pulling their parachute. They are unmoving when they leap out of their plane. Neglecting air resistance, what is the skydiver’s velocity when they pull the chute? How much work must be done by the parachute to slow the skydiver to 1 m s−1 at the ground level

Answer 1

Answer:

a) 420.2 meters per second

b) --6179846.4 joules

Explanation:

Using conservation of energy:

$K_i+U_i=K_f+U_f$

with Ki the initial kinetic energy, Kf the final kinetic energy, Ui the initial potential energy and Uf the final potential energy.

$\frac{mV_i^2} {2}+mgh_i=\frac{mV_f^2} {2}+mgh_f$ (2)

With Vi the initial velocity, Vf the final velocity, m the mass of the skydiver , g the gravitational acceleration, hi the initial altitude and hf the final altitude

Solving (2) for Vf:

$Vf=\sqrt{\frac{2}{m}(\frac{mV_i^2} {2}+mgh_i-mgh_f)}$

Vi is zero because they are unmoving when they leap out of their plane, hi is 10km and hf is 1 km:

$Vf=\sqrt{\frac{2}{70}((70)(9.81)(10000)-(70)(9.81)(1000))}=420.2 \frac{m}{s}$

By work-energy theorem:

$W=\Delta K=K_g-K_f$

with W the total work and ΔK the change on kinetic energy and Kg kinetic energy at ground level

$W=K_g-K_f=\frac{mV_g^2}{2}-\frac{mV_f^2}{2}$

$W=\frac{(70)(1)^2}{2}-\frac{(70)(420.2)^2}{2}=-6179846.4 J$