A baseball of mass 0.25 kg is hit so hard that it bounces on the ground and winds up in a spectator’s seat. If the ball leaves the ground at a speed 8.2 m/s and lands in the seat with a speed of 2.3 m/s, at what height, in metres relative to the ground, is the spectator?

Answer 1

The height of the spectator relative to the ground can be calculated using the equation:

h = (vf^2 - vi^2) / (2 * g)

where h is the height, vf is the final velocity, vi is the initial velocity, and g is the acceleration due to gravity (9.8 m/s^2).

Plugging in the given values, we get:

h = (2.3^2 - 8.2^2) / (2 * 9.8)

h = -7.2 m

Since the height is negative, this means that the spectator is 7.2 m below the ground.