Solve Ball's Acceleration: Vectors, 35m/s, 30m/s, 0.005s

[ahmed1237]

A tennis ball crossing a net at 35m/s [S10 degrees W] strikes a person's racket and her volley returns it at 30m/s [N 30 degrees W]. If the person's racket was in contact with the ball for 0.005s then what is acceleration of the ball?

For this question is my vi=35m/s and vf=30m/s? If so, when I draw my vectors do I subtract them or add them. The equation for acceleration is a=vf-vi/t right?

Please help and post your final answer and solution as I need to compare with mine!

 

[Mark44]

A tennis ball crossing a net at 35m/s [S10 degrees W] strikes a person's racket and her volley returns it at 30m/s [N 30 degrees W]. If the person's racket was in contact with the ball for 0.005s then what is acceleration of the ball?

For this question is my vi=35m/s and vf=30m/s? If so, when I draw my vectors do I subtract them or add them. The equation for acceleration is a=vf-vi/t right?

The ball's velocity is in the opposite direction after it is struck by the racket.

(v_f - v_i)/t -- notice the parentheses - is the change in velocity. That's not the same as the acceleration, although over short time intervals, the change in velocity divided by the change in time is closer to the instantaneous velocity. In other words, if Δt is reasonably small, Δv/Δt ≈ dv/dt = a.

Please help and post your final answer and solution as I need to compare with mine!

No, the way it works here is that you show us what you got, and we'll let you know if it's correct or not.

 

[ahmed1237]

so my answer is 4600m/s2 (w 21 degrees n) is that correct?

 

[Mark44]

The units should be m²/sec², but the acceleration value seems way high to me.

 

[ahmed1237]

so then how do I go about doing this?