# Help finding new velocity?

A full-sized glider has a weight of 4,900 N, while it's pilot has a weight of 825 N.

a. If it is 1.0 x 10^3 meters off the ground, how much potential energy do the plane and pilot have?

b. The same glider has a velocity of 35 m/s. How much kinetic energy does it have?

c. The same glider descends 9.0 x 10^2 meters. What is its new velocity?

I would like some help please. Especially with c. Even just an answer with no work. I got 140 m/s and I want to know if that is correct.

Thanks!

Relevance

First of all it is misleading to use the example of a glider in this question. A glider is known, by its very nature, take advantage of wind currents. So it picks up KE from the available wind. However, this question is presumably a "Free-Fall" question where wind will have no effect.

In order to give "a more natural feel" to this question why not use a stone or bowling ball in place of a "glider".

In short, I dislike very much question/s that confuse the reader's mind about nature. It is possible to use the models of 1st year physics while minimizing this type of confusion. Thus, the important principles of 1st year physics can be taught without making it seem so unreal.

Also, the word "velocity" stands for a vector quantity of physics that has BOTH size and direction . So I would think the more applicable quantity being asked about in this question = "speed" {ref part c.}

GPE = mgh = (4900+825)(9.81)(1000) = 56,162,250 ≈ 5.62E7 J ANS a.

KE = 1/2mV² = (0.5)(4900+825)(35)² = 0.5(5725)(35)² = 3,506562.5 ≈ 3.51E6 J ANS b.

ΔKE = mgh = (5725)(9.81)(900) = 50,546,025 ≈ 50.5E6 J

total KE at descended height = (50.5 + 3.5)E6 = 54.0E6

V = √2(KE)/m = √[2(54.0)E6/5725] = √0.018865E6 = 0.137E3 = 137 m/s ANS c.

.