## kinematics

Ronson drives a rocket sled from rest m up from a15 degree ramp at an acceleration of 8.0 m/s2. The rocket cuts off at the end of the ramp, which happens to be atthe edge of a 75 m high cliff. He freefalls in his sled until he bounces on a trampoline, which is on a 30.0 m high platform, and gives him an upward acceleration of 108 m/s2 for 0.45 s. Again he freefalls, bouncing this time on the ground, which gives him an upward acceleration of 445 m/s2 for 0.12 s. FInally after a third freefall, he stops bouncing, losing all vertical velocity but none of his horizontal velocity. Now that he's on the ground, he slides horizontally, decelerating at 1.5 m/s2 before coming to a stop. How far is the diagonal distance from the top of the ramp to his final resting place?

THis is really advanced, and I am really bad at projectiles motion. I know it involves x and y components.

It involves kinematic equations
V = Vo + at

X - Xo = Vot + .5at2

v2 = vo2 + 2a(X - Xo)

X - Xo = .5(Vo + V)t

If you can provide ideason how for me to solve this, that would be appreciated!

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 Mentor Blog Entries: 10 Welcome to PF. You need to take this one step at a time. Each step is simple, but there are many of them. Start with the ramp. You didn't say how long it is, but the problem statement must have given another piece of information that is missing from your post. Or is the ramp 75m high, i.e. the low end of the ramp is even with the bottom of the cliff?

## kinematics

sry
its supposed to be
"Ronson drives a rocket sled from rest 25 m up from a15 degree ramp at an acceleration of 8.0 m/s2."

 Mentor Blog Entries: 10 Okay, so what is Ronson's velocity when he reaches the end of the ramp?
 That's what I am trying to figure out with the different x and y components. Do you have any idea how to solve?

 Tags acceleration, distance, kinematics, velocity