Question

What is the maximum height of a ball with mass 500g thrown vertically upwards from the top of a roof of a building which height is 100m with a velocity of 20m/s^2?

Answer 1

To find the maximum height reached by the ball, we can use the following equations of motion:

v^2 = u^2 + 2as

Where: v = final velocity (0 m/s at the highest point) u = initial velocity (20 m/s) a = acceleration (acceleration due to gravity, approximately -9.8 m/s^2) s = displacement (maximum height reached, to be determined)

Plugging in the given values, the equation becomes:

0^2 = (20 m/s)^2 + 2*(-9.8 m/s^2) * s

Simplifying the equation:

0 = 400 m^2/s^2 - 19.6 m/s^2 * s

19.6 m/s^2 * s = 400 m^2/s^2

s = 400 m^2/s^2 / 19.6 m/s^2

s ≈ 20.41 meters

Therefore, the maximum height reached by the ball is approximately 20.41 meters.