Question

A ball hits a wall horizontally at 6.0 m/s. It rebounds horizontally at 4.4 m/s. The ball is in contact with the wall for 0.040 s.. The acceleration of the ball is:

-26 m/s sq.

-260 m/s sq.

-206 m/s sq.

206 m/s sq

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Answer 1

Given :

• Initial velocity (u) = 6.0 m/s
• Final velocity (v) = - 4.4 m/s (as it rebounds)
• Time (t) = 0.040 seconds

To find :

• Acceleration (a)

According to the question,

In this question we will use Newton's first equation of motion,

⇝ v = u + at

Where,

• v stands for Final velocity
• u stand for Initial velocity
• a stands for Acceleration
• t stands for Time

⇝ Substituting the values,

⇝ - 4.4 = 6.0 + a × 0.040

⇝ - 4.4 - 6.0 = 0.040a

⇝ - 10.4 = 0.040a

⇝ - 10.4 ÷ 0.040 = a

⇝ -260 = a

.°. The acceleration is -260 m/s².

So,the correct option is (b) - 260 m/s sq.

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