Question

PLEASE HELP ILL MARK BRAINLIEST!!! A ball is initially thrown downwards with an initial speed of 20 m/s from the top of a 300 m tall building. The ball has a constant downward acceleration of 10 m/s ^ 2 a). What is the velocity of the ball immediately before it hits the ground at the base of the building?

Answer 1

Using the 3rd equation of motion:

= v² - u² = 2gs ------ [g = Acceleration due to gravity]

= v² - 20² = 2 × 10 × 300

= v² - 400 = 6000

= v² = 6000 - 400

= v = √5600

= v = 74.83 m/s

And yeah it's done :)