Question

A paratrooper is initially falling downward at a speed of 27.6 m/s before her parachute opens. When it opens, she experiences an upward instantaneous acceleration of 74 m/s^2. (a) If this acceleration remained constant, how much time would be required to reduce the paratrooper's speed to a safe 4.95 m/s? (Actually the acceleration is not constant in this case, but the equations of constant acceleration provide an easy estimate.) (b) How far does the paratrooper fall during this time interval?

Answer 1

Answer:

a) 0.31 s

b) 19.77 m

Explanation:

We will need the following two formulas:

$V_{f} = V_{0}+at\\\\X=V_{0}t + \frac{at^{2}}{2}$

We first use the final velocity formula to find the time that it takes to decelerate the paratrooper:

$4.95\frac{m}{s}=27.6\frac{m}{s}-74\frac{m}{s^{2}}t\\\\-22.65\frac{m}{s}=-74\frac{m}{s^{2}}t\\\\t= \frac{22.65\frac{m}{s}}{74\frac{m}{s^{2}}}=0.31s$

Now that we have the time, we can use the distance formula to calculate the distance travelled by the paratrooper:

$X=27.6\frac{m}{s}*0.31s - \frac{74\frac{m}{s^{2}}*(0.31 s)^{2}}{2}=19.77 m$