1. Jul 21, 2011

### jehan4141

I have a physics problem with the solution provided, but the solution simply answers "No" with no numbers to back-up and explain why the answer is "No".

I have attempted the problem myself, but would like to see if my numbers are correct. Could you please tell me if my numbers are correct or incorrect?

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You can find the problem here on page 14 problem VII:

http://www.physics.princeton.edu/~mcdonald/examples/ph101_2006/learning_guide_ph101_2006.pdf

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**It is pretty important to look at the figure on this problem** :)

A stunt driver wishes to determine if he can drive off the incline and land on a platform (See the figure).

1. On the incline, the stunt driver can go from 0.0 m/s to 10.0 m/s in 2.0 seconds. Assuming constant acceleration, what speed can he achieve by the end of the incline if he starts at the bottom with speed v = 0.0 m/s?

Answer: 13.2 m/s ---> I understand this problem entirely. The next one was more difficult for me.

2. Show whether this speed is sufficient for him to make it.

I came to this conclusion by finding that it takes 1.312 seconds to cross the 15 m in the x-direction, when the car is in the air. Using t = 1.312 seconds, I found the height of the car (the height above the point of take-off) at that moment to be 0.224 m. Thus, 0.224 m < 2m.

Thus the car will not make it. My ultimate question: are these figures correct to explain why the speed is insufficient to make the jump safely?

Sorry this is so incredibly long :/ but I would truly appreciate any help you guys can send my way!!!

2. Jul 21, 2011

### Staff: Mentor

You might want to keep a few more decimal places in your intermediate results. Round only for final results that are to be submitted as an answer. What's your launch velocity to three decimal places?

Can you detail your calculation for the horizontal traversal time and the y-displacement at "impact"?

3. Jul 21, 2011

### jehan4141

Yes, of course!:) I will show my work with more decimal places.

From the first question, we found that the final velocity was v = 13.16074013 m/s.

I took this velocity to be the launch velocity. Since we know that the angle of the incline is 30 degrees:

Vox = Vx = 13.1607cos(30)

Using this Vx, we can find the time it took to travel the 15 meters in the horizontal direction.

Vx = x/t

t = x/Vx
t = 15/13.1607cos(30)
t = 1.3161 s

So it takes 1.3161 seconds to cross the 15 meters, in the horizontal direction.

I used this time (t=1.3161 seconds) to find the height above the point of take-off at this moment in time.

y = (Voy)t + 0.5at^2
y = 13.1607(1.3161) + 0.5(-9.8)(1.3161^2)
y = 0.19886 meters at impact. Thus, it will run into the platform wall!

***WOW - rounding makes such a difference!!! Thank you for the heads up on that. :)

So, the height above the point of take off is ~0.19886 meters. However, the height of the platform on which we wish to land on is 2 meters higher than the point of take-off.

0.19886 meters < 2 meters

Thus the car won't make it and the initial speed is insufficient!

Last edited: Jul 21, 2011
4. Jul 21, 2011

### Staff: Mentor

Nicely laid out work

Only one quibble: The value that you've used for the initial y-velocity: Voy. You've used the magnitude of the velocity vector rather than the y-component.

5. Jul 21, 2011

### jehan4141

My hero!!!! Thank you so much for pointing out my fault! Ahh I'm so happy!! :)

Ok, so after I correct this I should get:

Voy = 13.1607sin(30)
Voy = 6.58035 m/s

y = (Voy)t + 0.5at^2
y = (6.58035)(1.3161) + (0.5)(-9.8)(1.3161^2)
y = 0.17301 meters at impact. Thus, it will run into the platform wall!

0.17301 meters < 2 meters

Thus it will be insufficient to land on the other side!

6. Jul 21, 2011

### jehan4141

You are the bomb.com!!!!!!!!!!!!!!!!!!!!!!!!!! thank you sososososo much!!!!!

7. Jul 21, 2011

### Staff: Mentor

Yup.

I commend you for your clearly presented work!

8. Jul 21, 2011