Help With Kinetic and Potential Energy Problem?

Question

1) At what height h above the ground does the projectile have a speed of 0.5v?

Express your answer in terms of v and the magnitude of the acceleration of gravity, g.

2) What is the speed u of the object at the height of (1/2)h_max?

Express your answer in terms of v and g. Use three significant figures in the numeric coefficient.

*Note: I've already calculated and found h_max to be (v^2 / 2g)

Answer 1

well, let the speed at ground be v

by the conservation of energy

total initial energy = total final energy

total initial energy

considering the ground to be the h=0 for potential energy

is = 1/2 mv^2 (i.e. total KE)

total final energy will be

1/2 mV' ^2 + mgh

according to ques

V' = 1/2 v

therefore

1/2mv^2 = 1/2mV'^2 + mgh

cancell all m both sides and  *2

we get

v^2 = 0.25v^2 + 2gh

0.75v^2 = 2gh

h=(0.75v^2)/2g

ans 2)

since total initial energy = total final energy

therefore

1/2 m(Vinitial)^2 = mg(Hmax)  

so

   (1/2mV^2=mgH)  => mgH = mg(H/2) + 1/2mv^2



  mgH/2 = 1/2 mv^2

therefore v = sqrt(gH)

Answer 2

I would like to assume that "v" is the initial velocity of the body and based on your given data, the object was projected vertically upwards.

For part 1,your working formula is

Vf - v^2 = 2gh

where

Vf = 0.5v (given)

v = initial velocity (as assumed above)

g = acceleration due to gravity (value of which is constant but depends on whether you are using the English or Metric system)

h = height at which the velocity is 0.5v

Substituting values,

(0.5v)^2 - v^2 = 2(-g)(h)

NOTE the negative sign attached to the acceleration due to gravity. This simply implies that the body tends to slow down as it is going up.

Hence,

0.025v^2 - v^2 = -2hg

-0.975v^2 = -2gh

and solving for "h",

h = (0.975/2g)

h = 0.4875/g

**************************************...

Part 2

To determine hmax, use the same formula as above,

Vf^2 - v^2 = 2(g)(hmax)

and note that Vf = 0 at hmax.

Solving for hmax

0 - v^2 = 2(-g)(hmax)

Again, NOTE the negative sign attached to the accelaration due to gravity.

Solving for hmax,

hmax = v^2/2g and therefore,

(1/2)hmax = (1/2)(v^2/2g) = v^2/4g

Again, use the same formula to determine the speed of the body at hmax,

Vf^2 - v^2 = 2gh

where

Vf = velocity at (1/2)hmax

h = (1/2)hmax

and all the other terms are as previously defined.

Therefore,

Vf^2 - v^2 = 2(-g)(v^2/4g)

Vf^2 = (-v^2/2) + v^2

Vf^2 = v^2/2

Vf = v/sqrt 2

Vf = 0.707v