Solving a Difficult Projectile Motion Problem: How Far is the Diagonal Distance?

[Jessicaelleig]

Ronson drives a rocket sled from rest 25 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?

I know you find the diagonal distance by the Pythagorean theorem. One of the lengths of the sides is 75 m while you find the other by adding up the x distances of the jumps. I just don't know what numbers to use for velocity or acceleration in the beginning. Can someone please explain to me how to do this problem?

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!

 

[LowlyPion]

Ronson drives a rocket sled from rest 25 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?

I know you find the diagonal distance by the Pythagorean theorem. One of the lengths of the sides is 75 m while you find the other by adding up the x distances of the jumps. I just don't know what numbers to use for velocity or acceleration in the beginning. Can someone please explain to me how to do this problem?

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!

Just divide the problem up into multiple problems and figure out the answers applying the answer from one to the next phase of the problem.

1. Initial velocity off ramp? Figure acceleration and determine speed (Hint determine both x&y components based on angle)

2. Determine Velocity hitting trampoline. Consider effects of rebound and determine new Velocity.

3. Projectile motion now with new initial velocity off trampoline.

... and so on and so on and so on.

Since the presumption is that the sled doesn't lose horizontal velocity, then to figure the distance until it comes to rest you want to figure out all the time to accomplish these things and then multiply that by the horizontal velocity component.

 

[Jessicaelleig]

thank you :)

 

[LowlyPion]

thank you :)

I just realized that there was a horizontal deceleration phase, so total time until the deceleration and then the distance over the deceleration is what you need there.

Good luck.