Minimum Length of Emergency Escape Ramp for Truck to Stop

[seraphimhouse]

Homework Statement

In Fig. 8-36, a runaway truck with failed brakes is moving downgrade at 130 km/h just before the driver steers the truck up a frictionless emergency escape ramp with an inclination of θ = 15°. The truck's mass is 1.2 x 104 kg. What minimum length L must the ramp have if the truck is to stop (momentarily) along it?

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c08/fig08_35.gif

Homework Equations

Mechanical Energy
E,mec = K + U
K = 1/2mV^2
U = mgy

The Attempt at a Solution

Going through the problem I got K2 + U2 = K1 + U1 [2 being at the top of the inclined ramp and 1 being the bottom of the inclined ramp]. With substitution, I get:

1/2mv2^2 + mgy2 = 1/2mv1^2 + mgy1

Since velocity on the top of the ramp will equal zero and the length at the bottom of the ramp equals zero. the new formula will be:

gy2 = 1/2v1^2

subbing y2 with Lcos(theta):

Lcos(theta) = V1^2 / 2g

L = V1^2 / (2gcos(theta) )

I get an incorrect answer (247.97 m) The correct answer should be 260 m. Must be an misinterpretation somewhere. Any help would be nice :]

 

[Dr.D]

You don't know how to solve a right triangle by trig. Try again on that part. Most of your set up is fine.

 

[seraphimhouse]

You don't know how to solve a right triangle by trig. Try again on that part. Most of your set up is fine.

eff me. sin(theta) no cos(theta) thanks.