Answer to Question #147481 in Mechanics | Relativity for Jason Hirtle

Question

When a skydiver jumps out of an airplane she initially accelerates towards the earth at 9.81m/s squared. Wind resistance increases on the skydiver as they fall faster and faster through the air and eventually the skydiver reaches terminal velocity at 55.6m/s. When the diver opens her parachute, she experiences an acceleration of 30.0m/s squared. If the velocity must be 5.0 m/s at landing to avoid serious injury, then the minimum height above the ground that she must open her parachute would be ? __________ M

Answer 1

$v=v_0-at\implies t= \frac{v_0-v}{a},$

$h=v_0t-\frac{at^2}{2}=\frac{v_0^2-v^2}{2a}=\frac{55.6^2-5^2}{2 \cdot 30}=51 ~m.$