A 52.4 kg pole vaulter running at 10.8 m/s vaults over the bar.
The acceleration of gravity is 9.81 m/s2 .
If the vaulter’s horizontal component of ve- locity over the bar is 0.60 m/s and air resis- tance is disregarded, how high is the jump?
Answer in units of m.
Ke original=PE at top +ke at top
1/2 m v^2=mgh +1/2 m (.6)^2 solve for h
To determine the height of the jump, we can use the principles of projectile motion. We need to find the vertical component of the vaulter's initial velocity when they reach the peak of their jump.
First, let's determine the initial vertical velocity (Viy) of the vaulter using the given values. We know that the horizontal component of velocity (Vix) is 0.60 m/s, and the total velocity (Vt) is 10.8 m/s. We can use the Pythagorean theorem to find the vertical component of velocity:
Vt^2 = Vix^2 + Viy^2
Substituting the known values:
(10.8 m/s)^2 = (0.60 m/s)^2 + Viy^2
Vi^2 = (10.8 m/s)^2 - (0.60 m/s)2
Vi^2 = 116.64 m/s^2 - 0.36 m/s^2
Vi^2 = 116.28 m/s^2
Vi = √(116.28 m/s^2)
Vi ≈ 10.78 m/s
Now we can use the formula for vertical displacement (Δy) in projectile motion to find the height of the jump. The formula is:
Δy = (Vi^2 - Vf^2) / (2g)
Where:
Vi = Initial vertical velocity (10.78 m/s)
Vf = Final vertical velocity (0 m/s, at the peak)
g = Acceleration due to gravity (9.81 m/s^2)
Plugging in the values:
Δy = (10.78 m/s)^2 - (0 m/s)^2 / (2 * 9.81 m/s^2)
Δy = 116.4284 m^2/s^2 / 19.62 m/s^2
Δy ≈ 5.94 m
Therefore, the height of the vaulter's jump is approximately 5.94 meters.