Question

A ball is gently dropped from a height of 1 m (100 cm). Find its velocity and time after covering i) 50 cm ii) 100 cm distance.
( Acceleration a = g = 10 m/s2)

Answer 1

Here, Distance (s) = 100 cm

Initial velocity (u) = 0

Final velocity after covering 50 cm and 100 cm (v) = ? , Time after covering50 cm and 100 cm (t) = ?

acceleration (a=g) = 10 m/s2

After covering 50 cm velocity is:

According to the Newton's kinematic equation

v2 = u2 + 2as

v2 = 0 + 2×10×50

v2 = 0 + 1000

v2 = 1000

v = root 1000

v = 10 root 10

After covering 100 cm , velocity is :

According to the Newton's kinematic equation

v2 = u2 + 2as

v2 = 0 + 2×10×100

v2 = 2000

v2 = 10 root 20

After covering 50 cm , time is :

According to the Newton's kinematic equation

s = ut + 1/2 at2

50 = 0 + 1/2×10×t2

t2 = 50×2/10

t2 = 100/10

t2 = 10

t = root 10

After covering 100 cm , time is :

According to the Newton's kinematic equation

s = ut + 1/2 at2

100 = 0 + 1/2×10×t2

t2 = 100×2/10

t2 = 200/10

t2 = 20

t = root 20

I think so this is the answer