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Range and Elevation Equations

by r_swayze
Tags: elevation, equations, range
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r_swayze
#1
Sep8-09, 10:49 PM
P: 66
I have no idea where to begin with this problem. To me it seems like its missing information.

You desperately want to qualify for the Olympics in the long jump, so you decide to hold the qualifying event on the moon of your choice. You need to jump 7.52 m to qualify. The maximum speed at which you can run at any location is 5.90 m/s. What is the magnitude of the maximum rate of freefall acceleration the moon can have for you to achieve your dream?

any help?
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rl.bhat
#2
Sep8-09, 11:10 PM
HW Helper
P: 4,433
Quote Quote by r_swayze View Post
I have no idea where to begin with this problem. To me it seems like its missing information.

You desperately want to qualify for the Olympics in the long jump, so you decide to hold the qualifying event on the moon of your choice. You need to jump 7.52 m to qualify. The maximum speed at which you can run at any location is 5.90 m/s. What is the magnitude of the maximum rate of freefall acceleration the moon can have for you to achieve your dream?

any help?
Problem is based on the projectile motion. Can you state the relevant equations?
Here the range and the initial velocity is given.
What should the the angle projection for maximum range?
r_swayze
#3
Sep8-09, 11:34 PM
P: 66
Quote Quote by rl.bhat View Post
Problem is based on the projectile motion. Can you state the relevant equations?
Here the range and the initial velocity is given.
What should the the angle projection for maximum range?
according to the book the relevant equation is:

change in x = (-v^2 sin(2*theta)) / ay

I dont know theta so I dont think I can use this equation right?

and isnt 5.90 m/s the velocity of the x component? or is that the initial velocity of the jump?

Phrak
#4
Sep8-09, 11:47 PM
P: 4,512
Range and Elevation Equations

Quote Quote by r_swayze View Post
I dont know theta so I dont think I can use this equation right?
That's correct. Without calculus you can't calculate what trajectory will give you the greatest distance. On the other hand, maybe you were told it was 45 degrees.
rl.bhat
#5
Sep8-09, 11:49 PM
HW Helper
P: 4,433
Quote Quote by r_swayze View Post
according to the book the relevant equation is:

change in x = (-v^2 sin(2*theta)) / ay

I dont know theta so I dont think I can use this equation right?

and isnt 5.90 m/s the velocity of the x component? or is that the initial velocity of the jump?
The velocity is the initial velocity of jump. For maximum range, sin2θ should be 1 or θ should be 45 degrees.
Phrak
#6
Sep8-09, 11:56 PM
P: 4,512
How is theta defined in that equation?
rl.bhat
#7
Sep9-09, 12:04 AM
HW Helper
P: 4,433
Quote Quote by Phrak View Post
How is theta defined in that equation?
The angle through which the long jumper leaves the ground. But you have to assume him as a point object.


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