How High Can You Jump on the Moon and Sun Compared to Earth?

[gigi9]

Someone please help me how to do these problems below. Thanks a lot for your help.
1) On the surface of the moon the acceleration due to gravity is approximately 1/6 the sun at the surface of the earth, and on the surface of the sun it is approximately 29 times as great as at the surface of the earth. If a person on Earth can jump with enough initial velocity to rise 5ft, how high wil the same initail velocity carry that person (a) on the moon? (b) on the sun?
***I had s= -16t^2+ int. v*t+ int. s
s=5, int.s=0...what should I find, how do I do this problem?

2) Find the natural length of a spring if the work done in stretching it from a length of 2ft to the length of 3ft is one-fourth the work done in stretching it from 3ft to 5ft.
**Force=k*x,k is a constants, what I did was the force of stretching from 2ft-3ft is F1=k*1ft, F2=4*F1...not sure if I started out right...please show me how to do it.

3) A great conical mound of height h is built by the slaves of an oriental monarch, to commemorate a victory over the barbarians. If the slaves simply heap up uniform material found at ground level, and if the total weight of the finished mound is M, show that the work they do is 1/2h*M
**I'm totally stuck w/ this problem..what integral should i use?

 

[M98Ranger]

1) To solve this problem, we can use the equations of motion for a projectile: s=ut+1/2at^2 where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

(a) On the moon, the acceleration due to gravity is 1/6 of that on Earth, so we can substitute a=1/6g and solve for t when s=5ft and u is the initial velocity of the jump:
5ft = ut + 1/2(1/6g)t^2
Simplifying, we get:
5ft = ut + 1/12gt^2
Since we know that the initial velocity on the moon is the same as on Earth, we can substitute u=5ft/s and solve for t:
5ft = (5ft/s)t + 1/12gt^2
t = 10s
This means that it would take 10 seconds for the person to reach a height of 5ft on the moon.

(b) On the sun, the acceleration due to gravity is 29 times that on Earth, so we can substitute a=29g and solve for t when s=5ft and u is the initial velocity:
5ft = ut + 1/2(29g)t^2
Simplifying, we get:
5ft = ut + 14.5gt^2
Substituting u=5ft/s, we get:
5ft = (5ft/s)t + 14.5gt^2
t = 0.34s
This means that it would only take 0.34 seconds for the person to reach a height of 5ft on the sun.

2) In this problem, we can use the equation for work done: W=Fd where W is the work, F is the force, and d is the displacement.

We are given that the work done in stretching the spring from 2ft to 3ft is one-fourth the work done in stretching it from 3ft to 5ft. This means that:
W1 = 1/4W2
We also know that the force is directly proportional to the displacement, so we can write:
F1 = kx1
F2 = kx2
Substituting these into the equation for work, we get:
W1 = kx1d1