Question

A ball of mass 250 g is moving vertically upward with
initial velocity 12 m/s. A frictional force 0.25 N is constantly acts on the ball. What is the
highest position attained by the ball in upward direction.

Answer 1

we know that during upward motion,

v²-u² = 2as

where, v = final velocity = 0

u = initial velocity = 12m/s

s = total upward distance covered

a = net deceleration of the object

= deceleration due to gravity + deceleration offered by friction force

= 9.8 + (0.25/0.25).........................(acceleration = force / mass)

= 9.8 + 1

= 10.8

it will be taken -ve as it is acting downward

Therefore,

0² - 12² = 2 x (-10.8) x s

=> s = 144/21.6

=6.67 m

Hence the highest position attained by the object is 6.67 m

Answer 2

Given:

Mass = 250 g

Initial velocity = 12 m /s

Frictional force = 0.25 N

To find:

The  highest position attained by the ball in upward direction.

Solution:

By laws of motion,

v^2 - u^2 = 2 a s

Where,

v - final velocity

u - initial velocity

s - displacement

a - acceleration / deceleration

Here,

final velocity = 0

Deceleration

To find deceleration,

Deceleration = Gravity + Frictional force / Frictional force ( As the ball decelerates)

Hence,

9.8 + ( 0.25 / 0.25 )

Deceleration = -10.8 m / s

As it is moving in the reverse direction a negative sign is introduced.

Therefore,

Substituting the values,

0^2 - 12^2 = 2 x ( -10.8 ) x displacement

Displacement = 144 / 21.6

Displacement = 6.6 m in the upper direction.

Hence, the highest position attained by the object in the upward direction  is 6.6 m.