# Homework Help: Linear and projectile equation

1. Sep 25, 2007

### Remulak

1. The problem statement, all variables and given/known data

Jimmy is sitting in a 500kg car at the start of a quarter mile track, with a ramp at the end. His car also has two rockets strapped to the side of it. They each produce a 4000N force, and will burn until Jimmy cuts them off at the end of a quarter mile, and goes up the 10m high 25 degree ramp. Neglect air resistance and friction from tires.

2. Relevant equations
pretty much every linear and projectile equation
f = ma

3. The attempt at a solution
1/4 mile = 402m
m = 500kg, force = 8000n F = ma acceleration = 16 m/s^2
Plug 16m/s^2 into x-xo= vt -1/2at^2, solve for t = 7.08s, vf=vo + at, solve for final velocity gives 113.28m/s, when onto the ramp his velocities are 113.28sin25 and 113.28cos25 for the x and y components. When at the top of the ramp, use vf^2 = vo^2 + 2a(delta y) which gives 45.77 m/s for initial velocity in y direction. For max height, vyf = vyo - gt set vf = 0 gives 4.67s to reach peak height. y - yo = vt - 1/2gt^2, gives 106 m. Does this all look right? and how do I solve for the total distance?

Last edited: Sep 25, 2007
2. Sep 25, 2007

### Bryan52803

I'm assuming the problem is implying that the final velocity you derived (which appears correct) is the final velocity at the edge of the ramp in the appropriate directional vector, as you also have assumed.

The x-component of his velocity vector remains unchanged right? The only force is gravity which affects the y-component. So find out how long he's in the air, and find out how far in the x-direction he can move in that time.

Be careful about what you consider the ground to be...

3. Sep 25, 2007

### Remulak

would it be 14.22 seconds for him to come back to the ground? using a quadratic i found it was 9.55 seconds for him to reach 10m again from which he started. Then 0 = 45.77 - gt gives 4.67 seconds to finish the trip. 9.55 + 4.67 = 14.22 seconds.

Last edited: Sep 25, 2007
4. Sep 25, 2007

### Astronuc

Staff Emeritus
At the top of the ramp, one has a standard problem of a projectile with initial velocity at a given angle, starting at some elevation (10 m) about the final elevation.

On the ramp make sure the acceleration (or deceleration) takes into account the angle.

For reference - http://hyperphysics.phy-astr.gsu.edu/hbase/traj [Broken]

Taking off from the top of the ramp - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra9

Last edited by a moderator: May 3, 2017
5. Sep 25, 2007

### Remulak

do my values look correct? thanks for the link by the way. I got 1,460 m for his distance traveled after the ramp and 106m for the height.

6. Sep 25, 2007

### Remulak

I know how to do projectile motion questions but I'm confused on how to find the velocities at the top of the ramp because the car already has a velocity going up the ramp.

7. Sep 25, 2007

### Astronuc

Staff Emeritus
The car has a kinetic energy at the beginning of the ramp, and decreases its KE as it increases its gravitational PE (mgh).

Or one could calculate the V at the beginning of the ramp and the use the deceleration up the ramp to determine the V at the end of the ramp. The acceleration is simply for gravitational force component pointing down the ramp divided by the mass of the car.

The length of the ramp can be found by the height 10 m and the sin of the angle above the horizontal.