Projectile race track physics help

AI Thread Summary
The discussion revolves around solving a physics problem related to projectile motion for a racing event. The rider jumps off a slope at a 30° angle from a height of 1 meter and is airborne for 1 second. Key points include determining the speed at which the rider leaves the slope, the horizontal distance traveled before hitting the ground, and the maximum height reached. Participants clarify that the 1-second airborne duration indicates the total time in the air, not the time to reach the peak height. The use of the kinematic equation y=yi + vi t + 0.5 a t^2 is emphasized to find the vertical component of initial velocity.
Sam Fred
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Homework Statement


The Track for this racing event was designed so that riders jump off the slope at 30°, from a
height of 1m. During a race it was observed that the rider shown in Fig. 2 remained in mid air for
1s. Determine,
1) the speed at which he was traveling off the slope, ( 2points)
2) the horizontal distance he travels before striking the ground, (2points)
3) the maximum height above the ground he attains (3points)

Homework Equations


v=vi + at
y=yi + vi t + 0.5 a t^2

The Attempt at a Solution


in the attachment .
 

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hey man, welcome to physicsforums! When they say he remained in mid-air for 1 second, I think this means that he was airborne (not touching ground) for 1 second. Not that he was half-way through his jump at 1 second.
 


aha ... thanks man ... but how would i know the speed off he slope ?
Do i repeat the same steps with substituting t = 0.5 sec
 


No. That would work if his jump started and finished at the same height. But in this problem, it doesn't, because he starts the jump at 1m height, and finishes the jump at zero metres. So at t=0.5 sec, he is not at the highest part of the jump (he has already gone past that point). So the time when he has zero velocity will be some value, which is greater than 0.5 seconds.

You can use this equation you wrote: y=yi + vi t + 0.5 a t^2 (also, remember that vi needs to be the vertical component of initial velocity). You know the total change in height and you know the change in time, so you can simply solve for vi
 
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