Help with Vertical Motion Problem

  • Thread starter Thread starter theintarnets
  • Start date Start date
  • Tags Tags
    Motion Vertical
AI Thread Summary
A wrench dropped from rest reaches a speed of 24 m/s upon hitting the ground, leading to calculations for height and time of fall. Initial calculations yielded a time of 4.88 seconds and a height of 117 meters, but corrections indicated the height should be 29 meters and time approximately 2.4 seconds using the appropriate kinematic equations. The discussion also covers graphing height, velocity, and acceleration, emphasizing that velocity varies linearly with time while height varies with the square of time. The acceleration remains constant at 9.8 m/s², which should be reflected in the acceleration vs. time graph. The conversation highlights the importance of using correct formulas and assumptions, such as neglecting air resistance, for accurate results.
theintarnets
Messages
64
Reaction score
0

Homework Statement


I just want to know if I am doing this correct please!

A wrench is dropped from rest and hits the ground with a speed of 24 m/s. Find the height from which it was dropped and the length of time it took to reach the ground.

Homework Equations




The Attempt at a Solution


I used the formulas v(t) = vo - gt and h(t) = 1/2 gt2
So 24 m/s = 0 - 4.9t, and I get t = 4.88 seconds
And then h = 1/2(9.81)(4.882) and I got h = 117 meters


Also, the problem asks me to sketch graphs of height vs time, velocity vs time, and acceleration vs time. I have no idea how to do that, can someone help me understand how?
 
Physics news on Phys.org
Hello theintarnets,
You could have used v(t)=v(o)+gt(Your method indicates a negative time.
But still with proper sign the answer is correct and so is the formula somehow.)
The answer seems all fine to me provided you use correct value of g.(24 m/s = 0 - 4.9t -should be 24m/s=0+9.81gt h will get 1/4 th of its value)
For the graphs you can notice in the formulas that
velocity varies linearly with time and height varies as the square of time .Try plotting their graphs now.Be careful with the graph of h though because it asks you the height from the ground .
regards
Yukoel
[EDITED]
 
Last edited:
Your answer is incorrect. Try using the equation V2final=V2initial+2ax to find the height.
 
Last edited:
Whoops, thanks! I caught my mistake. I used this formula for vertical motion:
vy2 = voy2 + 2g(Δy)

I got 29 m for the height and then 2.4 seconds for the time. I've got the velocity vs time graph, but I have absolutely no idea how to make the acceleration vs time graph. The height graph is a bit fuzzy to me as well...
 
Nevermind, I figured out the height vs time graph, but I reeaalllyy need help with the acceleration graph please!
 
Every second it's speeding up by 9.8 m/s
 
It is being assumed that there is no air resistance and so the acceleration is that of free fall. Hence the acceleration is constant and that is what the graph of acceleration against time must show.
 
Just a thought, but if you wanted to try a different method you could think about the gravitational potential energy getting converted to kinetic energy as the wrench falls. This should give you an equation that you can rearrange for initial height.

As you might expect, you'll get exactly the same result when you rearrange the energy equation as when you rearrange Vf2 = Vi2+2ax. I don't think this helps much with the graphing though.
 
Solve for height using mgh=(.5)mv^2
Then use kinematics to solve for time.
 
Back
Top