Atennis ball with a speed of 24.4 m/s is moving perpendicular to a wall. after striking the wall, the ball rebounds in the opposite direction with a speed of 19.7884 m/s. if the ball is in contact with the wall for 0.0145 s, what is the average acceleration of the ball while it is in contact with the wall? take "toward the wall" to be the positive direction. answer in units of m/s² .

The acceleration of the ball is

Explanation:

We have given that ball strikes the wall at a speed of 24.4 m/sec

And after striking its goes in opposite direction with a speed of 19.7884 m/sec

We have given that speed towards the wall is positive

So u = 24.4 m/sec

And v = -19.7884 m/sec

Time is given as t = 0.0145 sec

We know that acceleration is given by

So the acceleration of the ball is

F = 91.2 kp

the correct answer is C

Explanation:

On a first degree lever, the fulcrum (pivot point) is between the weight to be lifted and the force applied. Knowing what force we must apply, write the endowment equilibrium equation.

We will assume that the counterclockwise direction of rotation is positive

             W d₁ - F d₂ = 0

              F = W d₁/ d₂

where

they indicate the value of the weight to lift W = 456 kp and its distance from the fulcrum d₁ = 36 cm; the distance to the applied force (F) is d₂ = 180 cm

let's calculate

            F = 456 36/180

            F = 91.2 kp

therefore the correct answer is C