Projectile Motion of ice skater

AI Thread Summary
The discussion focuses on calculating the starting point for an ice skater's stunt involving projectile motion off a ramp inclined at 15°. The skater must achieve a horizontal distance of 6.40 m while falling 3.80 m vertically. Initial calculations using the Pythagorean theorem indicated a hypotenuse of 7.44 m, but further analysis revealed that the skater needs to reach a speed of 2.97 m/s to cover the horizontal distance in 0.88 seconds. Participants suggest using energy analysis or force analysis to determine the necessary height on the ramp for the skater to achieve this speed. The conversation emphasizes the importance of understanding kinematic equations and the effects of gravitational acceleration on the skater's motion.
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Homework Statement



You are planning a stunt to be used in an ice skating show. For this stunt a skater will skate down a frictionless ice ramp that is inclined at an angle of 15° above the horizontal. At the bottom of the ramp, there is a short horizontal section that ends in an abrupt drop off. The skater is supposed to start from rest somewhere on the ramp, then skate off the horizontal section and fly through the air a horizontal distance of 6.40 m while falling vertically for 3.80 m, before landing smoothly on the ice. How far up the ramp should the skater start this stunt?

Homework Equations





The Attempt at a Solution



I found, using the Pythagorean theorem, that the distance diagonally (the hypotenuse) is 7.44m, but I am kind of stuck there.

Thanks a lot for you help!
Damian
 
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Start by finding how long it takes for an object to fall 3.8m vertically using gravitational accel. Then use that time to relate how fast she must be going to travel 6.4m in the time that you find. That is a start
 
so, using d= vt + (5.)(a)(t*t) <---- that's t squared, sorry :)

i did 3.8 = (0)(t) + (.5)(9.8)(t*t)

and found t to be be .88s

then i did

6.4 = v(.88) + (.5)(9.8)(.88)

and got v to be 2.97m/s

so that means that the skater will have to be going 2.97m/s. it this correct so far?
 
the "short horizontal section" in the problem implies that there will be no acceleration in the horizontal direction. So your second formula is wrong
 
so it would just be d = vt

6.4 = v(.88)

= 7.27 m/s ?

thanks
 
can someone explain where to go from here? thanks
 
Try using energy analysis (PE and KE) to find how far up she needs to be to achieve that speed...
 
im sorry, i don't think we have covered that yet in class. i tried to look it up online, but I am not really having any luck. is there any other way to solve?
 
Use force analysis to find how fast an object accelerates down a ramp that is 15 degrees. after that just use normal kinematic equations to relate distance/velocity/acceleration
 

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