Projectile motion motorcycle question

[kelvin56484984]

Homework Statement

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?
the answer is 355

Homework 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

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

 

[haruspex]

Vf=Vi*cosθ

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

 

[kelvin56484984]

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

Vx=Vi*cosθ

 

[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 traveled i.e. d=V_projectioncosθ. No of barrels will be simply the horizontal distance since the diameter of each barrel is 1m.

 

[haruspex]

Vx=Vi*cosθ

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

 

[kelvin56484984]

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

 

[cnh1995]

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

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.

 

[kelvin56484984]

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

 

[cnh1995]

Right! You have Vi already.

 

[kelvin56484984]

Vi equal to 60?

 

[cnh1995]

Vi equal to 60?

Yes.

 

[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?

 

[cnh1995]

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

 

[haruspex]

But how to use projectile velocity find the Vi?
50*sin30=Vi*sin30*t -1/2gt^2

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.