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1. Homework Statement
http://img132.imageshack.us/img132/2847/img1dy7.jpg [Broken] shows an 11.7 ft wide ditch with approach roads at an angle of 15 degrees.With what minimum speed should a motorbike be moving on a road so that it safely crosses the ditch.Assume the length of the bike to be 5 ft, and it leaves the road when the front part runs out of the approach road.
(in the figure, i've shown only 1 angle as 15 deg but both are 15)
2. Homework Equations
equations of motion
3. The Attempt at a Solution
Let the velocity be v
[tex]v_x[/tex]=vcos15
[tex]v_y[/tex]=vsin15
i've used the fact that the time the bike stays in air above the roads should be the time it sould take to cross the distance of the ditch
then,
0=[tex]v^2*sin15+2gh[/tex]
or h=(v^2*(sin15)^2)/2g
After this, [tex]v_y[/tex]=0, so the bike falls
using s=ut+0.5gt^2
i get,
t=vsin15/g
So, total time= 2vsin15/g
This the time it takes to cover horizontal dist
so,
vcos15*2vsin15/g=11.7
[tex]v^2[/tex]sin30/g=11.7
g=32.15 ft/s^2
solving i get v=27.42 ft/s
But in my book, the answer is given to be 32 ft/s
Can any1 plz post the error in this problem,or a simpler method ?
Thanks
PS: Sorry for my poor latex and drawing skills, but i think that i made myself clear :tongue:
http://img132.imageshack.us/img132/2847/img1dy7.jpg [Broken] shows an 11.7 ft wide ditch with approach roads at an angle of 15 degrees.With what minimum speed should a motorbike be moving on a road so that it safely crosses the ditch.Assume the length of the bike to be 5 ft, and it leaves the road when the front part runs out of the approach road.
(in the figure, i've shown only 1 angle as 15 deg but both are 15)
2. Homework Equations
equations of motion
3. The Attempt at a Solution
Let the velocity be v
[tex]v_x[/tex]=vcos15
[tex]v_y[/tex]=vsin15
i've used the fact that the time the bike stays in air above the roads should be the time it sould take to cross the distance of the ditch
then,
0=[tex]v^2*sin15+2gh[/tex]
or h=(v^2*(sin15)^2)/2g
After this, [tex]v_y[/tex]=0, so the bike falls
using s=ut+0.5gt^2
i get,
t=vsin15/g
So, total time= 2vsin15/g
This the time it takes to cover horizontal dist
so,
vcos15*2vsin15/g=11.7
[tex]v^2[/tex]sin30/g=11.7
g=32.15 ft/s^2
solving i get v=27.42 ft/s
But in my book, the answer is given to be 32 ft/s
Can any1 plz post the error in this problem,or a simpler method ?
Thanks
PS: Sorry for my poor latex and drawing skills, but i think that i made myself clear :tongue:
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