Kinematics with projectile motion(2 questions)

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To safely jump a 50 m wide gorge from a 1 m high ramp inclined at 30 degrees, Jim must reach a speed of 23 m/s, calculated using the projectile motion formula. Jake, on the other hand, attempts the jump with a 45-degree ramp at 20 m/s but will not make it, falling short by approximately 9 to 10 m. The calculations for both scenarios involve determining the horizontal and vertical components of velocity and the time of flight. The formula used for Jim's jump is 50 = (v²*sin(2*30))/9.81, while Jake's scenario requires solving a quadratic equation to find the time before he drops below ground level. Overall, both problems illustrate the application of kinematics in projectile motion.
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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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