Question

A ball hits a wall horizontally at 6m/s. It rebounds horizontally at 4.4m/s. The ball is in contact with the wall for 0.04s. What is the acceleration of the ball?

Answer 1

In this Case

Given

• Initial Velocity (u) = 6 m/s
• Final Velocity (v) = -4.4 m/s (as the direction of ball is opposite)
• Time (t) = 0.04s

To Find

• Acceleration

Calculating Acceleration

Formula Used :- v = u + at

Substituting Values

☞ -4.4 = 6 + a × 0.04

☞ -4.4 = 6 + 0.04a

☞ -4.4 - 6 = 0.04a

☞ -10.4 = 0.04a

☞ -104/10 = 004/100a

☞ -104 = 4/10a

☞ -104 × 10 = 4a

☞ -1040 = 4a

☞ a = -1040/4

☞ a = -260

Hence, we got Acceleration = -260 m/s²

Additional Information

• Rate of Change of Velocity with respect to time is known as Acceleration
• SI unit of Acceleration is m/s²
• Acceleration is a Vector Quantity

Three Equations of Motion

• v = u + at
• S = ut + ½at²
• v² - u² = 2as

Answer 2

Given :

• Initial velocity (u) = 6m/s
• Final velocity (v) = - 4.4 m/s
• Time (t) = 0.04s

To find :

• Acceleration of ball (a)

Solution :

According to first equation of motion

→ v = u + at

→ - 4.4 = 6 + a*0.04

→ - 4.4 = 6 + 0.04a

→ - 4.4 - 6 = 0.04a

→ - 10.4 = 0.04a

→ a = - 10.4/0.04

→ a = - 260 m/s²

Hence,

• Acceleration of the ball is - 260 m/s²

Additional Information :

• The rate of change in velocity is known as acceleration
• S.I unit of acceleration is m/s²
• Minus (-) shows retardation