ball thrown perpindicular, v1= 10.9 m/s, bounces off wall, v2=8.45 m/s, contact with wall for .0115 s, a?

Answer 1

Remember that velocity is a vector and has both a magnitude (speed) and direction component.

The ball’s initial velocity is 10.9 m/s toward the wall.
The ball’s final velocity is 8.45 m/s away from the ball.
By bouncing off the wall, the direction the ball was traveling reversed. We can indicate this numerically by calling the direction toward the wall as being “positive” and the direction away from the wall as being “negative”. We can thus write the initial (v1) and final (v2) velocities as,
V1 = +10.9 m/s,
V2 = -8.45 m/s.

The change in velocity (delta v) in this case is the final velocity minus the initial velocity,
(delta v) = v2 – v1
(delta v) = -8.45 m/s – 10.9 m/s
(delta v) = -19.35 m/s

Acceleration is a change in velocity with respect to time.
Acceleration = change in velocity / change in time

We have previously calculated the change velocity (delta v = -19.35 m/s), and we are told that the ball is on contact with the wall for .0115 seconds…this is our change in time value, (delta t).

Acceleration = (delta v) / (delta t)
Acceleration = (-19.35 m/s) / (.0115 s)
Acceleration = -1683 m/s^2

Notice that the acceleration is negative and remember what we said earlier about a negative sign (referring to velocity at the time). We previously defined the direction away from the wall as being the “negative” direction. Since acceleration, like velocity, is a vector…it has both a magnitude and direction.
The magnitude of the acceleration is 1683 m/s^2 and is pointed away from the wall.