Maximum force exerted by floor?

[kristibella]

1. A 220 g ball is dropped from a height of 2.2 m, bounces on a hard floor, and rebounds to a height of 1.7 m. The figure shows the impulse received from the floor.

What maximum force does the floor exert on the ball?

Homework Equations

To find the velocity, I used vi=\sqrt{}2gh and vf=\sqrt{}2gh. To find the impulse, I used F= \stackrel{}{}m(vf-vi)t

The Attempt at a Solution

Vi = 6.57 m/s (I used 2.2m for h)
Vf = 5.77 m/s (I used 1.7 m for h)
t= .005 s
F = \stackrel{(220)(5.77-6.57)}{.005} = -35,200

 

[LowlyPion]

1. A 220 g ball is dropped from a height of 2.2 m, bounces on a hard floor, and rebounds to a height of 1.7 m. The figure shows the impulse received from the floor.

What maximum force does the floor exert on the ball?

Homework Equations

To find the velocity, I used
v_i = \sqrt{2gh}
and
v_f = \sqrt{2gh}.
To find the impulse, I used
F= \frac{m(vf-vi)}{\Delta t}

The Attempt at a Solution

Vi = 6.57 m/s (I used 2.2m for h)
Vf = 5.77 m/s (I used 1.7 m for h)
t= .005 s
F = \stackrel{(220)(5.77-6.57)}{.005} = -35,200

Mind your units
F_{avg} = \frac {(.220)(5.77-6.57)}{.005} = -35.2N This is the Average Force.

Your Force Graph indicates that it is Triangular, hence the Maximum Force is twice the Average.

F_max = 70.4 N

 

[Nositi]

Mind your units
F_{avg} = \frac {(.220)(5.77-6.57)}{.005} = -35.2N This is the Average Force.

Your Force Graph indicates that it is Triangular, hence the Maximum Force is twice the Average.

F_max = 70.4 N

I found this post while googling my problem just to see how others had attempted it, and I noticed another mistake here. You have to consider the direction of the momentum. One of the velocities should be negative, because they are going in opposite directions. Therefore we have 5.77 + 6.57 for J_x rather than 5.77 - 6.57. F_max should actually be 1085.92 N here.