quick physics question

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Semidevil

Diamond Member
Apr 26, 2002
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Assume that from a running start you are able to jump straight up in the air, go up a height h = 0.86 m, and come back down at exactly the same spot where you started. Neglecting air resistance, what is the velocity with which you took off from the ground?

can someone walk me through how to do this?? How can I solve it with only one piece of info?
 
V

Viper GTS

Lifer
Oct 13, 1999
38,107
433
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I'll give you a hint:

When you left the ground you had kinetic energy equal to the potential energy at the point you peaked.

Viper GTS
 
S

Semidevil

Diamond Member
Apr 26, 2002
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can't find the edit button, but here is question # 2.

You have been hired as the technical advisor on an action movie. In one stunt in the film, a truck needs to jump a ditch that is d =9.5 m wide. The side of the ditch on which the truck lands is a height h = 0.9 m lower than the side from which it takes off. The sides are level so the truck takes off horizontally.
What is the minimum initial velocity necessary for the truck to make the jump?


 
V

Viper GTS

Lifer
Oct 13, 1999
38,107
433
136
Originally posted by: Semidevil
can't find the edit button, but here is question # 2.

You have been hired as the technical advisor on an action movie. In one stunt in the film, a truck needs to jump a ditch that is d =9.5 m wide. The side of the ditch on which the truck lands is a height h = 0.9 m lower than the side from which it takes off. The sides are level so the truck takes off horizontally.
What is the minimum initial velocity necessary for the truck to make the jump?
Click to expand...

For that, you need to find a velocity such that the truck travels at least it's length (axle to axle) + 9.5m in the time it would take it to drop 0.9m.

Viper GTS
 
Legendary

Legendary

Diamond Member
Jan 22, 2002
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Originally posted by: Semidevil
Assume that from a running start you are able to jump straight up in the air, go up a height h = 0.86 m, and come back down at exactly the same spot where you started. Neglecting air resistance, what is the velocity with which you took off from the ground?

can someone walk me through how to do this?? How can I solve it with only one piece of info?
Click to expand...

PE = mgh
KE = 1/2mv2

mgh = 1/2mv2
(9.8)(.86) = .5(v2)

solve for v, i'm too lazy. v2 = v squared

and viper i believe with the caliber of question 1 the length of the truck is not taken into account. also because it's not given.

.9 = .5at2
1.8=9.8t2
t = sqrt(1.8/9.8)

d = vt
9.5 = (v) (t (taken from above))
solve for v
enjoy.
 
V

Viper GTS

Lifer
Oct 13, 1999
38,107
433
136
Originally posted by: Semidevil
I still dont know how to do #1..............:-(
Click to expand...

At any given time you have kinetic and potential energy.

Assume when you leave the ground you have potential energy 0, and kinetic energy k.

k = ½mv²

You want to know v.

When you peak, you're at a height of 0.86m. You now have kinetic energy of 0 (since you stop briefly at the peak) but potential energy k (no energy is lost, it just changed form). Figure out your potential energy at that moment (just use m for your mass, since it will cancel in the end), then use the formula above to determine v.

Viper GTS
 
YetioDoom

YetioDoom

Platinum Member
Dec 12, 2001
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Originally posted by: TheEvil1
you need more then that to solve that you need some time or an acceleration
Click to expand...



You have an acceleration - gravity.

-9.8 m/s^2

 
Y

Yossarian

Lifer
Dec 26, 2000
18,010
1
81
Another way to think about it is, what would the final velocity be of an object dropped from that height? With no air resistance, it is the same as the velocity required to propel an object to that height in the first place.

distance = 0.5at^2, solve for t
then use

distance = velocity/t, solve for v
 
T

Templeton

Senior member
Oct 9, 1999
467
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I'm assuming that you haven't gone over energy yet, kinematics is usually taught first, and if you you mastered that, you could figure it out. So here's how to do question 2 using basic kinematics equations:

S = Vot + 1/2 at^2

Sx = Voxt
9.5m = Voxt

Sy = 1/2 at^2
.9m = 1/2(9.8 m/s^2) t^2
Solve for t
t^2 = (.9m) / (4.9 m/s^2)
t = sqrt(.18367 s^2)

plug that value into the equation for the x axis:
9.5m = Vox(sqrt(.18367 s^2))
solve for Vox, should work out to 22.2 m/s
Hope this helps, good luck
 
H

Haircut

Platinum Member
Apr 23, 2000
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Originally posted by: Templeton
I'm assuming that you haven't gone over energy yet, kinematics is usually taught first, and if you you mastered that, you could figure it out. So here's how to do question 2 using basic kinematics equations:

S = Vot + 1/2 at^2

Sx = Voxt
9.5m = Voxt

Sy = 1/2 at^2
.9m = 1/2(9.8 m/s^2) t^2
Solve for t
t^2 = (.9m) / (4.9 m/s^2)
t = sqrt(.18367 s^2)

plug that value into the equation for the x axis:
9.5m = Vox(sqrt(.18367 s^2))
solve for Vox, should work out to 22.2 m/s
Hope this helps, good luck
Click to expand...

I get 4.11 m/s as the answer.

Vf^2 - Vi^2 = 2*a*s from the kinematics equation.
Vf=0, a=-9.8 m/s^2

so, Vi^2 = 2*9.8*0.86, Vi^2 = 16.86, Vi = 4.11 m/s

The agrees with the value which we get if we use an energy calculation:

1/2 * m *v^2 = m * g * h.
Cancel m from both sides, g=9.8 m/s^2

v^2 = 2 *9.8 * 0.86 as before.
 
T

Templeton

Senior member
Oct 9, 1999
467
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Haircut:
That answer doesn't seem to work
You're dropping a distance of .9m, you have to make it to the other side at the time that it takes to drop that distance. using .9m =1/2(9.8 m/s^2) t^2 , the time to drop that distance must be .429 seconds, at 4.11 m/s you could only make it 1.76m before dropping .9m in the y.

Edit:
I just realized that you were solving question1, I answered the second question asked in the thread. I agree that question 1 is 4.11m/s , sorry for the confusion.
 
H

Haircut

Platinum Member
Apr 23, 2000
2,248
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Originally posted by: Templeton
Haircut:
That answer doesn't seem to work
You're dropping a distance of .9m, you have to make it to the other side at the time that it takes to drop that distance. using .9m =1/2(9.8 m/s^2) t^2 , the time to drop that distance must be .429 seconds, at 4.11 m/s you could only make it 1.76m before dropping .9m in the y.
Click to expand...
Oops, I didn't see you were doing question 2 :eek:
Your answer to that one is correct.
 
N

NogginBoink

Diamond Member
Feb 17, 2002
5,322
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Originally posted by: Semidevil
can't find the edit button, but here is question # 2.

You have been hired as the technical advisor on an action movie. In one stunt in the film, a truck needs to jump a ditch that is d =9.5 m wide. The side of the ditch on which the truck lands is a height h = 0.9 m lower than the side from which it takes off. The sides are level so the truck takes off horizontally.
What is the minimum initial velocity necessary for the truck to make the jump?
Click to expand...

So the truck is going to descend 0.9 m.

Using formula for acceleration and acceleration due to gravity, find the time it'll take to fall 0.9 m.

Then find the speed at which the truck will travel a horizontal distance of 9.5 meters in that period of time.
 
S

Semidevil

Diamond Member
Apr 26, 2002
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wow, thanx guys. you guys are smart. Now, here is a couple more that I would like explained. I already have the answers, but I just dont really know where it came from.

you are driving on a road at 100km/hr and hit your brakes. The brakes accelerate at a rate of -.5 m/s^2. how far will you travel after hitting the brakes before you stop?

ok, so I have 100 km/hr. That is the velocity right? Can I just call that the average velocity? and I have -.5 as acceleration. so a = -.5.

the teacher did this to solve it:

0 = at + v(o).

so
t = -v(o)/a; which is (27.8m/s)/ (5m/s). so t = 5.56s

first question. where did the 27.8 come from??

then he just plugged in the equation and the answer came up to be 77.3m.

 
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