Answer to Question #290204 in Mechanics | Relativity for appi

Question

A parachutist of mass 70 kg jumps from a plane at an altitude of 32 km above the surface of the Earth. Unfortunately, the parachute fails to open. (In the followingparts, neglect horizontal motion and assume that the initial velocity is zero.)

(a) Calculate the time of fall (accurate to 1 s) until ground impact, given no air resistance and a constant value of g.

(b) Calculate the time of fall (accurate to 1 s) until ground impact, given constant g and a force of air resistance given by

Answer 1

Explanation & Calculations

• From the start to the end, you can apply the 4 motion formulae appropriately as the acceleration (gravitational acceleration in this case) is constant through out.
• Before that, some given quantities need to be converted into SI units which then can be substituted into the formulae to calculate the needed quantities.

$\qquad\qquad \begin{aligned} \small 32\,km &=\small32000\,m\\ \end{aligned}$

• Then you can proceed,

$\qquad\qquad \begin{aligned} \small \downarrow s&=\small ut+\frac{1}{2}at^2\\ \small 32000\,m&=\small 0+\frac{1}{2}\times9.8\,ms^{-2}\times t^2\\ \small t&=\small 80.8\\ &\approx\small 81\,s \end{aligned}$

• For the second case (here some data is missing) if the air resistance is given as a fixed force, then you can calculate the fixed acceleration at which he falls down.
• $\qquad\qquad \begin{aligned} \small\downarrow F&=\small ma\\ \small mg-f&=\small ma_1\\ \small a_1&=\small \frac{70\times9.8-f}{70}\\ \end{aligned}$
• Then, apply the same formula just as above to calculate the new time.