Magnitude of average acceleration

[matthew1991]

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

 

[diazona]

OK, well to start you off: the problem asks for an acceleration. What equations do you know that might allow you to find acceleration?

 

[matthew1991]

a= (v final - v initial) / (t final - t initial)
a = dv/dt
a d^2(s)/d(t)^2

 

[matthew1991]

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)

 

[rwisz]

.I can provide some guidance on how to approach this problem. First, we need to understand the concept of acceleration. Acceleration is the rate of change of velocity over time. In simpler terms, it measures how quickly an object's velocity is changing.

In this case, the golf ball is released from a height of 1.50 m and bounces back to a height of 0.62 m. This means that the ball's velocity changes from a positive value (upward) to a negative value (downward). The time it takes for this change to occur is given as 5.39 ms.

Using the formula provided in the second statement, we can calculate the average acceleration of the golf ball while it is in contact with the floor. Since the ball is in contact with the floor for the entire 5.39 ms, we can use this time as the value for "time" in the formula. The initial velocity (v(initial)) is 0 m/s since the ball is released from rest.

Substituting these values into the formula, we get:
a = (0.62 m/s - 0 m/s) / (5.39 x 10^-3 s)
a = 0.62 m/s / 5.39 x 10^-3 s
a = 115.2 m/s^2

Therefore, the magnitude of the average acceleration of the golf ball while it is in contact with the floor is 115.2 m/s^2. This value is negative because the ball is accelerating downwards due to gravity.

I hope this explanation helps you understand how to approach this problem. If you have any further questions, please do not hesitate to ask.