Kinematics with projectile motion(2 questions)

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Discussion Overview

The discussion revolves around two physics problems related to projectile motion involving a skateboard jump across a gorge. Participants explore the calculations needed to determine the required speed for a successful jump and analyze the implications of different ramp angles and speeds.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Jim needs to achieve a speed of 23 m/s to successfully jump a 50 m wide gorge from a ramp inclined at 30 degrees, but the method to arrive at this speed is unclear to some participants.
  • Jake believes he can jump the gorge with a ramp inclined at 45 degrees and a speed of 20 m/s, but it is questioned whether he will make it across, with some suggesting he will fall short by 9 m.
  • One participant suggests calculating the horizontal and vertical components of velocity to determine the time until the jumper reaches ground level, which could help solve both problems.
  • Another participant provides a formula for calculating the jump distance, indicating that the height of the ramp must be considered in the calculations.
  • There is a mention of a quick-fix approach to the calculations, which some participants believe may overlook important factors, such as the ramp height.

Areas of Agreement / Disagreement

Participants express uncertainty about the calculations and methods used to determine the jump distances and speeds. While some calculations are proposed, there is no consensus on the correctness of the approaches or the final answers.

Contextual Notes

Participants note the importance of considering the height of the ramp in the calculations, and there are unresolved mathematical steps in determining the time of flight and horizontal distance traveled.

koolaid123
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1) Jim decides to jump a gorge on his skateboard. The gorge is 50 m wide and jim has constructed a 1 m high ramp iclined at 30 degrees to the horizontal. If air resistance is neglected, calculate what speed Bart must reach if he is to land safely on the other side of the gorge.

The answer is 23 m/s but I don't know how to get it.

2) Jake has calculated that he can jump the jorge if the ramp is inclied to 45 degrees, and he attains a speed of 20 m/s. Will Jake make it? If not how much short of the other side is he when he drops below the level of the ground?

The answer is no, and 9 m but i don't know how to get it.
 
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Looks like homework.

With v as initial velocity, can you calculate the horizontal and vertical component of the velocity?
Based on the vertical component, can you calculate the time until Jim/Bart is at floor level again?
Based on that time, can you calculate the horizontal position at that point?

That can be used for both 1 and 2.
 
koolaid123 said:
1) Jim decides to jump a gorge on his skateboard. The gorge is 50 m wide and jim has constructed a 1 m high ramp iclined at 30 degrees to the horizontal. If air resistance is neglected, calculate what speed Bart must reach if he is to land safely on the other side of the gorge.

The answer is 23 m/s but I don't know how to get it.

2) Jake has calculated that he can jump the jorge if the ramp is inclied to 45 degrees, and he attains a speed of 20 m/s. Will Jake make it? If not how much short of the other side is he when he drops below the level of the ground?

The answer is no, and 9 m but i don't know how to get it.

1) Use this
50 = (v2*sin(2*30))/9.81
v=23.xx m/sec

2) Jake doesn't make it. Misses by 9m to 10m. You can use the same formula above but more accurate is
-1 = (20*sin45)t -4.91*t2
(-1, because the ramp is 1 m high). Solve for t. Use value of t in
x = (20*cos45)*t.
x will be much lower than 50m.

I think I'm right, but may be not.
 
Neandethal00 said:
1) Use this
50 = (v2*sin(2*30))/9.81
v=23.xx m/sec
I think you forgot the height of the ramp here.
 
mfb said:
I think you forgot the height of the ramp here.

Didn't forget, tried a quick-fix. ha ha
It gives the same answer OP was looking.
Thanks.
 

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