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Magnitude of average acceleration

  • #1
1. A golf ball released from a height of 1.50 m above a concrete floor, bounces back to a height of 0.62 m. If the ball is in contact with the floor for 5.39 ms, what is the magnitude of the average acceleration a of the ball while it is in contact with the floor?



2. v=v(initial)+acceleration*time
acceleration = -9.8 m/s^2



]3. I don't know where to start =|. Please assist
 

Answers and Replies

  • #2
diazona
Homework Helper
2,175
6
OK, well to start you off: the problem asks for an acceleration. What equations do you know that might allow you to find acceleration?
 
  • #3
a= (v final - v initial) / (t final - t initial)
a = dv/dt
a d^2(s)/d(t)^2
 
  • #4
Problem solved. 1.65*10E3 m/s^s.
For all who may come along seeking help: initial velocity is the velocity when the ball first contacts the floor and final velocity is the velocity at which the ball leaves. Delta T is the time in contact with the floor.

With the distanc equation you can solve for the time and then multiply by acceleration to find velocity.
0=initial position-1/2 a t^2 solve for t
plug t into v=at
this is initial velocity -5.42 m.s

final velocity ises a different equation
v final ^2 = v(initial)^2-2a(final p- initial p)
 

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