# Homework Help: Projectile motion motorcycle question

1. Feb 20, 2016

### kelvin56484984

1. The problem statement, all variables and given/known data
A motorcycle daredevil wants to ride up a 50.0m ramp set at a 30.0 θ incline to the ground. It will launch him in the air and he wants to come down so he just misses the last of a number of 1.00 m diameter barrels. If the speed at the instant when he leaves the ramp is 60.0 m/s, how many barrels can be used?

2. Relevant equations
Vf=Vi*cosθ
x=xo+vi*cos θ*t
Vf=Vi*sin θ -gt
y=y0+vi*sin θ*t-1/2gt^2
height=Vi^2*sin θ*cos θ/2g
Range=Vi^2*sin2 θ/g
3. The attempt at a solution
Vf=Vi*cos30
Vi=60 / cos30
Vi=69.3 m/s

R=Vi^2*sin2 θ/g
R=69.3^2*sin60/9.8
R=424

I

2. Feb 20, 2016

### haruspex

That's a new one on me. What do you mean by vi and vf there?

3. Feb 20, 2016

### kelvin56484984

I find it on my book
It should be
Vx=Vi+at for x component
(since a=0)

Vx=Vi*cosθ

4. Feb 20, 2016

### cnh1995

You need to sketch a diagram first. First calculate the time after which the rider will land. Use that time to find the horizontal distance travelled i.e. d=Vprojectioncosθ. No of barrels will be simply the horizontal distance since the diameter of each barrel is 1m.

5. Feb 20, 2016

### haruspex

That makes more sense. Which of those velocities is given?

6. Feb 20, 2016

### kelvin56484984

vx is equal to 60m/s?
I try to sketch this diagram

7. Feb 20, 2016

### cnh1995

Vx is not 60m/s. It is the total velocity of projection, inclined at 30 degrees with the horizontal. That's what you are supposed to show in your diagram.
Now, from your diagram, you can see the net vertical displacement of the rider. You know the acceleration due to gravity g. Set up an equation relating displacement with time and g and find the time taken by the rider to reach the ground. You can then use this time to find the horizontal distance covered before landing.

Last edited: Feb 20, 2016
8. Feb 20, 2016

### kelvin56484984

But how to use projectile velocity find the Vi?
50*sin30=Vi*sin30*t -1/2gt^2
x=x0+Vi*cos30t
these equation?

9. Feb 20, 2016

### cnh1995

10. Feb 20, 2016

### kelvin56484984

Vi equal to 60?

11. Feb 20, 2016

### cnh1995

Yes.

12. Feb 20, 2016

### kelvin56484984

But I substitute vi=60 to
50*sin30=Vi*sin30*t -1/2gt^2
I get t=0.995s or 5.127s
then put the t =5.127 into
x=x0+vi*cos30*t
then x=266
what's wrong with it?

13. Feb 20, 2016

### cnh1995

Well, the displacement should be negative i. e. -25m.

14. Feb 20, 2016

### haruspex

No. He leaves the ramp at a speed of 60m/s. At that time, he is already at a height of 50 sin(30) m.