A 56.1 kg pole vaulter running at 9.50 m/s vaults over the bar.
The acceleration of gravity is 9.81 m/s? If the vaulter's horizontal component of velocity over the bar is 0.70 m/s and air resistance is disregarded, how high is the jump?
To determine the height of the jump, we can use the principles of projectile motion. Let's break down the problem into two parts: the vertical and horizontal components.
First, let's calculate the time it takes for the vaulter to reach the highest point of their jump. We can use the vertical component of the vaulter's initial velocity.
Vertical component of velocity (v_y) = 9.50 m/s
Acceleration due to gravity (a) = 9.81 m/s²
Using the equation v = u + at (where v is final velocity, u is initial velocity, a is acceleration, and t is time), we can solve for time:
0 m/s (at the highest point) = v_y - g * t
0 = 9.50 - 9.81 * t
Solving for t, we get:
t = 9.50 / 9.81
Now that we have the time it takes for the vaulter to reach the highest point, we can calculate the maximum height of the jump.
Using the equation s = ut + 0.5at² (where s is the displacement, u is the initial velocity, t is time, and a is acceleration), the displacement at the highest point will be zero (as the vaulter will be momentarily stationary). Therefore:
0 = v_y * t + 0.5 * a * t²
Substituting the values we know:
0 = 9.50 * (9.50 / 9.81) + 0.5 * 9.81 * (9.50 / 9.81)²
Solving for the displacement (maximum height), we get the answer to the question.