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Help with Vertical Motion Problem 
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#1
Jun1412, 12:00 AM

P: 64

1. The problem statement, all variables and given/known data
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. 2. Relevant equations 3. The attempt at a solution I used the formulas v(t) = v_{o}  gt and h(t) = 1/2 gt^{2} So 24 m/s = 0  4.9t, and I get t = 4.88 seconds And then h = 1/2(9.81)(4.88^{2}) 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? 


#2
Jun1412, 12:17 AM

P: 73

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] 


#3
Jun1412, 12:25 AM

P: 235

Your answer is incorrect. Try using the equation V^{2}final=V^{2}initial+2ax to find the height.



#4
Jun1412, 01:26 AM

P: 64

Help with Vertical Motion Problem
Whoops, thanks! I caught my mistake. I used this formula for vertical motion:
v_{y}^{2} = v_{oy}^{2} + 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... 


#5
Jun1412, 01:50 AM

P: 64

Nevermind, I figured out the height vs time graph, but I reeaalllyy need help with the acceleration graph please!!



#6
Jun1412, 02:05 AM

P: 235

Every second it's speeding up by 9.8 m/s



#7
Jun1412, 02:48 AM

P: 950

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.



#8
Jun1412, 03:42 AM

P: 112

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 V_{f}^{2} = V_{i}^{2}+2ax. I don't think this helps much with the graphing though. 


#9
Jun1412, 08:58 AM

P: 48

Solve for height using mgh=(.5)mv^2
Then use kinematics to solve for time. 


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