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Question with gravity, speed, heights

  • Thread starter ital_dj
  • Start date
31
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[SOLVED] question with gravity, speed, heights

1. Homework Statement
a boy fires a 60g pebble with his slingshot. the pebble leaves the slingshot at 35m/s.
a) how high above the slingshot will the pebble rise if it is fired straight up?
b) if the pebble is fired so that it goes in an arc and has a speed of 10m/s at its maximum height, what will the maximum height be?
c) at what speed would an 80g pebble have to be fired to reach the same height as the pebble in (a)? Assume that the 80g pebble is also fired straight up.


2. Homework Equations
I'm just going to guess some relevant equations, because I'm so lost.
d = vt
F = mg
Eg = mg[tex]\Delta[/tex]h


3. The Attempt at a Solution
I tried basic multiplication, division, and other stuff... couldn't figure it out

You're help is greatly appreciated. Thanks a lot

EDIT: I just looked at the teacher's solution that she wrote, and it was mgh = [tex]\frac{mv_{1}^{2}}{2}[/tex], but how did she arrive at that? Thanks
Also, from that equation it gets narrowed down to h = [tex]\frac{v_1^2}{2g}[/tex]
 
Last edited:

Answers and Replies

161
0
are you sure you copied the solution correctly? remember that when the pebble is shot straight up and reaches its maximum height that 100% of its kinetic energy will have been converted into gravitational potential energy.
 
31
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Sorry, you're right... it should be divided by 2... I'll fix it right now. But, with the correct solution now, how did she arrive at that?
 
161
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what is the equation for kinetic energy? what is the equation for gravitational potential energy?
 
31
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what is the equation for kinetic energy? what is the equation for gravitational potential energy?
Kinetic Energy = 1/2mv[tex]^2[/tex]
Gravitational Energy (Eg) = mg[tex]\Delta[/tex]h
 
5
0
i guess max height which is required in prblm (a) = u^2/2g

u = initial velocity = 35

so max height = 61.25 m i think we don't need mass here o_O i'm confused too...
 

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