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

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)

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Answer 2

Given:

A ball is dropped on top of a building. The ball reaches the floor in 5 sec.

To find:

• Initial velocity
• Height of building
• Final velocity just before it touches ground.

Calculation:

• Since the question says that the ball is DROPPED, it means that initial velocity is zero ( i.e. u = 0 m/s).

$h = ut + \dfrac{1}{2} g {t}^{2}$

$\implies h = 0 + ( \dfrac{1}{2} \times 10 \times {5}^{2} )$

$\implies h = 125 \: m$

• So, height of building is 125 metres.

$v = u + at$

$\implies v = 0 + gt$

$\implies v = 0 + (10 \times 5)$

$\implies v = 50 \: m {s}^{ - 1}$

• So, final velocity is 50 m/s.