Question

A ball is dropped on top of a building. a) What is the inertial velocity of the ball? b) What is the height of the building if the ball reaches the floor in 5 s ? c) What is the velocity of the ball just before it hits the ground

Answer 1

Answer:

its easy

I will tell you next time

Answer 2

Answer:

Explanation:

Given : A ball is dropped on top of a building

the ball reaches the floor in 5 s

To Find : initial velocity of the ball

height of the building

velocity of the ball just before it hits the ground

Solution:

ball is dropped

Hence, initial velocity of the ball = 0 m/s

a = g = 10 m/s²

S = ut + (1/2)at²

u = 0

t = 5

a = g = 10

S = 0 + (1/2) * 10 * 5²

=> S = 125 m

Height of the building is 125 m

(122.5 m if g is taken 9.8 m/s² )

V = Velocity of the ball just before it hits the ground

V = u + at

=> V = 0 + 10(5)

=> V = 50 m/s

( 49 m/s if g is taken 9.8 m/s²)