Optimizing Motorbike Jump Over Slope: Solving for Minimum Speed

[f(x)]

Homework Statement

http://img132.imageshack.us/img132/2847/img1dy7.jpg 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)

Homework Equations

equations of motion

The Attempt at a Solution

Let the velocity be v
$$v_x$$=vcos15
$$v_y$$=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=$$v^2*sin15+2gh$$
or h=(v^2*(sin15)^2)/2g
After this, $$v_y$$=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
$$v^2$$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 please 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:

 

[f(x)]

got it finally
i used length to be 11.7, but should have taken it to be 16.7 ft
Still I am getting 32.7 ft/s but i think this much error is permissible, specially due to conversion of g in terms of ft/s^2

Anyways if any1 of u has a better soln., do put it up here

 

[andrevdh]

The first answer you got is the same when using the range equation

$$R = \frac{v_o ^2 \sin(\theta _o)}{g}$$

My approach to it is to assume that the jump is just successfull if the rear tyre reaches the edge of the bank at the opposite side. Assuming that the bike do not rotate during the jump the x-distance the front wheel covers will then be

$$11.7 + 5\cos(15^o)$$

from this the jump time will be

$$\frac{17.11}{v_o}$$

the front wheel will then be

$$5\sin(15^o)$$

up in the air at touch down on the opposite side. This then gives one a launch speed of 38.7 feet per second according to my calculations.