Question

during his recent skydiving adventure, luke autbeloe had reached a terminal speed of 14.5 m/s as he approached the ground with his parachute. during an attempt to snap one last photo with his camera, luke fumbled it from a height of 70.6 m above the ground.
a. determine the speed with which the camera hits the ground.
speed m/s
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b. determine the time for the camera to free fall from lukes hands to the ground.
time s
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1. k20q4

the zero gravity research facility at nasa - operated glenn research center in ohio is used to test the behavior of fluids, flames, equipment and other objects in free fall. it consists of a 467 - foot long, 12 - foot diameter, steel vacuum chamber. the steel chamber resides inside of a concrete lined shaft which extends 510 feet below ground level. objects falling through the tower experience free fall over a distance of 130 meters (426 feet) before being brought to a halt.
a. determine the falling time for objects dropped from rest.
time s
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b. determine the final speed of the objects before the braking period begins.
final speed m/s
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Explanation:

Step1: Identify the kinematic equation

For part a, use the equation \(v^{2}=v_{0}^{2}+2gh\). Here, \(v_{0} = 14.5\space m/s\), \(g = 9.8\space m/s^{2}\), and \(h=70.6\space m\).

$$v^{2}=(14.5)^{2}+2\times9.8\times70.6$$

$$v^{2}=210.25 + 1383.76$$

$$v^{2}=1594.01$$

$$v=\sqrt{1594.01}\approx39.9\space m/s$$

Step2: Use the kinematic equation for time

For part b, use the equation \(h = v_{0}t+\frac{1}{2}gt^{2}\). Substitute \(h = 70.6\space m\), \(v_{0}=14.5\space m/s\), and \(g = 9.8\space m/s^{2}\) into \(70.6=14.5t + 4.9t^{2}\). Rearrange to \(4.9t^{2}+14.5t - 70.6=0\). Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 4.9\), \(b = 14.5\), \(c=-70.6\).

$$t=\frac{-14.5\pm\sqrt{(14.5)^{2}-4\times4.9\times(-70.6)}}{2\times4.9}$$

$$t=\frac{-14.5\pm\sqrt{210.25 + 1383.76}}{9.8}$$

$$t=\frac{-14.5\pm\sqrt{1594.01}}{9.8}$$

Take the positive root \(t=\frac{-14.5 + 39.9}{9.8}=\frac{25.4}{9.8}\approx2.6\space s\)

Step3: For the NASA problem part a (falling from rest)

Use \(h=\frac{1}{2}gt^{2}\), with \(h = 130\space m\) and \(g = 9.8\space m/s^{2}\). Then \(t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times130}{9.8}}=\sqrt{\frac{260}{9.8}}\approx5.15\space s\)

Step4: For the NASA problem part b

Use \(v = gt\), with \(t\approx5.15\space s\) and \(g = 9.8\space m/s^{2}\). Then \(v=9.8\times5.15 = 50.5\space m/s\)

Answer:

a. \(39.9\space m/s\)
b. \(2.6\space s\)
NASA a. \(5.15\space s\)
NASA b. \(50.5\space m/s\)