Question

Mike clears a crossbar while pole vaulting. If Mike releases the pole before achieving peak height, and it takes 1 second to fall from the peak height to the landing pit (1m high), how high above the ground was Mike at the peak height of the vault?

a. 3m
b. 4m
c. 5m
d. 6m

Answer 1

Mike was approximately 4 meters above the ground at the peak height of the vault.

Step-by-step explanation:

To determine the height above the ground at the peak height of the vault, we can use the kinematic equation for free-fall motion:

h = 1/2 * g * t^2

Where h is the height above the ground, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes to fall from the peak height to the landing pit.

In this case, t = 1 second and h = 1 meter. Using these values, we can plug them into the equation:

1 = 1/2 * 9.8 * t^2

Simplifying the equation gives us:

t^2 = 0.2041

Taking the square root of both sides yields:

t = 0.4523 seconds

Therefore, Mike was at a height of approximately 4 meters above the ground at the peak height of the vault.