The pole vaulter runs 100m as fast as he can in 10 seconds. This gives him a velocity of 10m/s. The pole vaulter weighs 70kg. The bar is 5 m high. The pole vaulter jumps at an angle of 74 degrees. How long is he in the air? How far does he travel from the beginning of the jump to the end of the jump? How high did he get above the pole? Use the projectile motion formulas.

Vf= Vi + gt
change of y= Vit + .5gt^2
change of x= Vxt
Vfy^2= Viy^2 + 2g(change of y)

Question

The pole vaulter runs 100m as fast as he can in 10 seconds. This gives him a velocity of 10m/s. The pole vaulter weighs 70kg. The bar is 5 m high. The pole vaulter jumps at an angle of 74 degrees. How long is he in the air? How far does he travel from the beginning of the jump to the end of the jump? How high did he get above the pole? Use the projectile motion formulas.

Vf= Vi + gt
change of y= Vit + .5gt^2
change of x= Vxt
Vfy^2= Viy^2 + 2g(change of y)

anonymous · Answer

write where your up to and we can see what part your stuck on.

faf3 · Answer

i found Vx=10m/s and Vy=0m/s but thats just for the run to the pole before the jump. Im stuck with the actual jump

anonymous · Answer

so he jumps at 74 degrees
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anonymous · Answer

whats the initial velocity as he jumps in the y and x direction (Vy and Vx here) presume the 10 m/s is all converted in his jump at angle 74 degrees(the hypotenuse)

anonymous · Answer

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