an object launched vertically upward?

Question

let us consider an object launched vertically upward with an initial speed v. Neglect air resistance.

At what height h above the ground does the projectile have a speed of 0.5v?
Express your answer in terms of and the magnitude of the acceleration of gravity g.

ok so i found hmax to be (v^2/2g) and i thought that the h at 0.5v would be 1/4of hmax and it's telling me that i am off by a multiplicative factor.

Answer 1

This problem is an energy problem. Since we are ignoring air resistance, there are no non-conservative forces. Therefore this problem is independent of mass.... That's stated so often, why do people still deny it??? Anyway...\

Initially, h = 0 so PEi = 0. That's about it :P


Ei = Ef
½m v² = ½m (½ v)² + mg h

The m's cancel

½ v² = ½ (½ v)² + g h

½ v² = ½ * ¼ v² + gh

½ v² = ⅛ v² + gh

½v² - ⅛ v² = gh

(3/8) v² / g = h

Or..

3v² / (8g) = h

Answer 2

since you found max height already by doing:

0 = v^2 - 2g(maxheight)
maxheight = v^2 / 2g

using the formula:
v^2 = v0^2 + 2a(height)

(v/2)^2 = v^2 - 2g(height)
2g(height) = v^2 - (v/2)^2 = v^2 - (v^2 / 4)
height = (3v^2 / 4) / 2g

so, the height when the velocity is only 1/2 of the initial velocity is 3/4 of the max height.

Answer 3

Let us start out with the Conservation of Energy Theory
Eo=Ef
KEo = KEf+PEf
1/2mvo² = 1/2mvf² + mgh

If you notice, mass cancels out:

1/2vo² = 1/2vf² + gh

We are told vf = .5vo, so let us plug that in.

1/2vo² = 1/2 (.5Vo)² + gh
1/2vo² = .125Vo² + gh
1/2vo² - .125Vo² + gh
.375vo² = gh
h= (.375vo²)/g

This is the lowest you can simplify it without knowing vo.

Good Luck.

Answer 4

All you have to do is plug in the numbers for your variables. Square 0.5m/s and then divide by 2 time 9.8 and you have your answer.

Answer 5

it would dpend on the weight of the object