Question

A ball hits horizontally at 6.0ms-1. It rebounds horizontally at 4.4 ms-1. The ball is in contact with the wall for 0.04 s. What is the acceleration of the ball .





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Answer 1

Given:

• Initial velocity,u = 6.0 m/s

• Final velocity, v = -4.4 m/s ( Given : rebounds horizontally )

• Time taken = 0.04 s

To be calculated:

Calculate the acceleration of given ball ?

Formula used:

Acceleration = change in velocity/Time taken

Solution:

Acceleration = change in velocity/time taken

Acceleration = v - u / t

★ Substituting the values in the above formula,we get

Acceleration = -4.4 - 6.0 / 0.04

= - 10.4 / 0.04

= - 1040 / 4

= -260 m/s²

Thus, the acceleration of given ball is -260 m/s².

Answer 2

$\huge{\underline{\pink{\tt{Given,}}}}$

• A ball hits horizontally (Initial Velocity) = 6m/s (Forward)
• It rebounds horizontally (Final Velocity) = -4.4 m/s (Forward) And 4.4 m/s (backward)
• The Ball is contact with the wall (Time Taken ) = 0.04 seconds

$\huge{\underline{\pink{\tt{To \ Find,}}}}$

• The Acceleration of the Ball .

$\huge{\underline{\pink{\tt{Solution :}}}}$

$\longrightarrow$ We Know that :

$\bigstar \boxed{\sf{Accleration = \frac{Final\:Velocity - Initial\:Velocity}{time}}}$

(Always  Initial velocity and final velocities are opposite in direction )

Now Put the Value in this Formula :

$\implies \sf{Acceleration = \dfrac{(-4.4 - 6)}{0.04}}$

$\implies \sf{Acceleration = \dfrac{-10.4}{0.04}}$

$\implies \sf{Acceleration = \dfrac{-1040}{4}}$

$\implies \boxed{\sf{Acceleration = -260m/s^{2}}}$

Therefore,

$\longrightarrow$ Acceleration of the Ball = 260 m/s² (Answer)

$\rule{200}2$