Question

A skydiver stepped out of an airplane at an altitude of 1000m fell freely for 5.00s opened her parachute and slowed to 7.00m/s in a negligible time what was the total elapsed time from leaving the airplane to landing on the ground

Answer 1

Answer:

t = 17.68s

Explanation:

In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:

$y=y_o-v_ot-\frac{1}{2}gt^2$            (1)

y: height for a time t

yo: initial height = 1000m

vo: initial velocity = 0m/s

g: gravitational acceleration = 9.8m/s^2

t: time = 5.00 s

You replace the values of the parameters to get the values of the new height of the skydiver:

$y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m$

Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.

You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0

$0=877.5-7.00t-4.9t^2$       (2)

The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:

$t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s$

You use the positive value of t1 because it has physical meaning.

Finally, you sum the times of both parts of the trajectory:

total time = 5.00s + 12.68s = 17.68s

The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s