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Projectile Motion: Range

by gsr_4life
Tags: motion, projectile, range
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gsr_4life
#1
Jan16-06, 06:29 PM
P: 3
A motorcycle daredevil is attempting to jump across as many buses as possible. The takeoff ramp makes an angle of 18.0 degrees above the horizontal, and the landing ramp is identical to the takeoff ramp. The buses are parked side by side, and each bus is 2.74m wide. The cyclist leaves the ramp with a speed of 33.5m/s. What is the max number of buses the cyclist can jump?
I know that this is a question of range involving a vector x. I know that Vo =33.5m/s, Vox= 31.86m/s and Voy= 10.35m/s using sin and cos of 18degrees. What I am having trouble with is finding time "t" so I can use the range equation. I am not even sure if I am going about this right. Any help would be great, thanks.
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berkeman
#2
Jan16-06, 06:40 PM
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If you shoot a ball up into the air with the initial vertical velocity of Voy= 10.35m/s, how long before it comes back down to its starting height? (And what will its downward Vy be then?)
cscott
#3
Jan16-06, 07:24 PM
P: 786
Realize your projectile takes a parabolic path, therefore [tex]d = v_ot + 1/2at^2[/tex], where d = 0.

gsr_4life
#4
Jan17-06, 11:49 PM
P: 3
Projectile Motion: Range

I set up a table of values for displacement of y=0, a=-9.8, Vo=10.35 and used the equation y=Vot+1/2a * (t squared) and plugged t into the equation Voxt=R and divided it by 2.74 = 24 buses


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