Physics homework help projectile motion

[scratch1]

physics homework help please projectile motion

Homework Statement

a biker is iding on a flat stretch of lan 50 m long, without friction. he then goes up a flat ramp that its surface is 30 degrees from the ground (no friction) the biker then leaves the ramp and goes 37.5 m across to the other side, and lands on a flat surface that is level with the top of the ramp where the biker rode off of. what is the constant acceleration the biker needs to acclerate while riding through the 50 m flat stretch to successfully land on the other flat surface 37.5 away from the end of the ramp? (no acceleration at ramp)

Homework Equations

v final squared - v initial squared= 2ad (for beginning 50m)
displacement y= vt-4.9t squared
displacement x= vt

The Attempt at a Solution

v final squared - v initial squared= 2ad (for beginning 50m)
displacement y= vt-4.9t squared
displacement x= vt

i then said v leaving the ramp was v sin 30, so v= v sin 30 for the last two equations
but I am not sure when it leaves the ramp, should i call the speed v sin or v cos? or just v??

*this is my 1st time posting so srry if its confusing to read through this. thanks

 

[scratch1]

wait did i post in the wrong place?

 

[gneill]

What's the height or length of the ramp?

 

[scratch1]

its not given

 

[gneill]

its not given

Without the height or length of the ramp I think that the best you could do is provide a symbolic solution (with ramp length or ramp height as a variable).

If the ramp height is unknown, then the amount of kinetic energy the biker will lose while climbing the ramp is unknown, and so the velocity he needs to attain on approach to the ramp is unknown.

 

[zgozvrm]

By "no friction," I hope you're referring to air friction (otherwise known as air drag or air resistance). Otherwise, that biker isn't going anywhere!

 

[scratch1]

i think the bike remains at a constant velocity when it goes up the ramp. and my bad, i meant the only external forces are gravity and the force to accelerate the car in the beginning

 

[gneill]

It's getting complicated. Acceleration on a frictionless surface, gravity that doesn't work on ramps, and now a car is involved!

Perhaps the problem is poorly posed in the original source? I suppose we could assume a very short ramp...

 

[PhysDrew]

It's been a while since I've done these types of probs, but if we know the horizontal range of the 'projectile' (ie 37.5m), we could sub into;
R=[v initial squared*sin2*30]/g

where R is the horizontal range. Then solve for the inital velocity, and go from there...

 

[gneill]

It's been a while since I've done these types of probs, but if we know the horizontal range of the 'projectile' (ie 37.5m), we could sub into;
R=[v initial squared*sin2*30]/g

where R is the horizontal range. Then solve for the inital velocity, and go from there...

But there's nowhere to go from there. You need to know the velocity at the base of the ramp, but you don't know how much velocity will be lost climbing the ramp.

 

[PhysDrew]

But there's nowhere to go from there. You need to know the velocity at the base of the ramp, but you don't know how much velocity will be lost climbing the ramp.

Yeah I just assumed constant velocity up the ramp...

 

[scratch1]

im sorry i got this problem ona test a week ago, but i wasn't sure how to do it, so i tried to rewrite it to the best of my memory. but velocity is assumed to be constant going up the ramp. I am just no sure what would you consider the bikes speed to be when you leave the ramp, if you call the speed going to the ramp v. would you call that speed whe n you leave the ramp v cos, v sin, or just v?

 

[gneill]

im sorry i got this problem ona test a week ago, but i wasn't sure how to do it, so i tried to rewrite it to the best of my memory. but velocity is assumed to be constant going up the ramp. I am just no sure what would you consider the bikes speed to be when you leave the ramp, if you call the speed going to the ramp v. would you call that speed whe n you leave the ramp v cos, v sin, or just v?

Velocity in the x-direction would be v cos(θ). Velocity in the y-direction would be v sin(θ). θ is the angle of the ramp with respect to the horizontal. The speed of the bike remains v.

 

[scratch1]

ok thanks i get it now