# Max Force of Floor on Ball: Calculate

• brunettegurl
In summary, the floor exerts a maximum force of 444.4 newton (N) on the ball when it's dropped from a height of 2.60 m and bounces and rebounds.
brunettegurl
whats the max force??

## Homework Statement

A 275.0 g ball is dropped from a height of 2.60 m, bounces on a hard floor, and rebounds to a height of 1.71 m. The impulse received from the floor is shown below. What maximum force does the floor exert on the ball if it is exerted for 8.00 ms?

## Homework Equations

$$\Delta$$p = m1(v2-v1) and $$\Delta$$p = 0.5bh

## The Attempt at a Solution

since there's no given velocity i decided to take the differcence between the 2 heights and divide it by the time to get a speed i used the first equation from which i got a change in momentum and then tried to figure out the height since the base = time (8ms) is this reasoning correct??

#### Attachments

• kn-pic0928_new.png
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Use conservation of energy to find the velocity before impact and after impact.

can u pls. elaborate?? thanx

What's the final speed of a ball dropped 2.60 m? (Use kinematics or energy conservation to find that speed.) Similarly, what's the initial speed of a ball that rises to a height of 1.71 m?

Find the impluse, then compare it to the area under the curve in your diagram.

ok so my energy conservation wld look like :: (mgh1)+ (0.5mv^2)= (mgh2)+ (.5m[v2]^2)

but if i try to use a kinematic equation to find the v on the LHS of the equation I am missing part of the equation. would i have to half the time given to me to figure out the velocity for the LHS?

brunettegurl said:
ok so my energy conservation wld look like :: (mgh1)+ (0.5mv^2)= (mgh2)+ (.5m[v2]^2)
Energy is not conserved during the collision, otherwise the ball would just bounce back up to its starting height. But you can use energy conservation to find the speed of the ball just before it hits the ground: mgh = 0.5mv^2.

but if i try to use a kinematic equation to find the v on the LHS of the equation I am missing part of the equation. would i have to half the time given to me to figure out the velocity for the LHS?
The time given is the time for the collision with the floor--it has nothing to do with the kinematics of the falling (or rising) ball. Don't try to use it to find the speed. You'll need that time to find the peak force using impulse = ∫Fdt.

so i used mgh = 0.5mv^2 to find the speed @h = 2.60 and the speed @ h=1.71 m and put it into the equation impulse = Fdeltat... is that the right way of going abt it??

brunettegurl said:
so i used mgh = 0.5mv^2 to find the speed @h = 2.60 and the speed @ h=1.71 m and put it into the equation impulse = Fdeltat... is that the right way of going abt it??
Assuming you did it correctly, yes. Use the speeds to find the change in momentum (careful with signs); set that change in momentum equal to the impulse, which is ∫Fdt.

what do u mean by being careful with the signs?? wouldn't both of them be positive??

brunettegurl said:
what do u mean by being careful with the signs?? wouldn't both of them be positive??
No. Just before hitting the floor, the ball is going down; just after bouncing, it's going up.

ok so i compensated for that and I'm still not getting the right answer so I'm going to show you my wrkings incase i did something wrong in that

[when its going down] h= 2.60
mgh=.5mv^2
sqrt(2gh) = v
-7.139 m/s = v1
[when its going up] h= 1.71
mgh=.5mv^2
sqrt(2gh) = v
+5.789m/s = v2

then impulse = Fdt
imulse/dt = F
m(v2-V1)/dt = F
(0.275kg)(+5.789-(-7.139)) / 8x10^-3 s = F
therefore F= 444.4 N

but the comp. app is telling me that answer is wrong...so what did i do wrong??

brunettegurl said:
[when its going down] h= 2.60
mgh=.5mv^2
sqrt(2gh) = v
-7.139 m/s = v1
[when its going up] h= 1.71
mgh=.5mv^2
sqrt(2gh) = v
+5.789m/s = v2
This is good.

then impulse = Fdt
The impulse is not simply F*Δt, because the force isn't constant. Instead, it's ∫Fdt (that's an integral sign), which means you need the area under the F-t graph. (What's the area of a triangle?)

You're almost done.

thanx

## 1. What is the "Max Force of Floor on Ball" calculation used for?

The "Max Force of Floor on Ball" calculation is used to determine the maximum force that a floor exerts on a ball when it is dropped or thrown onto the floor.

## 2. How is the "Max Force of Floor on Ball" calculated?

The "Max Force of Floor on Ball" is calculated by multiplying the mass of the ball by the acceleration due to gravity (9.8 m/s²) and adding the impact force of the ball on the floor.

## 3. Why is it important to calculate the "Max Force of Floor on Ball"?

It is important to calculate the "Max Force of Floor on Ball" because it helps in understanding the impact of throwing or dropping a ball on the floor. It can also be used to determine the strength and durability of the floor and the ball.

## 4. What factors can affect the "Max Force of Floor on Ball"?

The "Max Force of Floor on Ball" can be affected by the mass and velocity of the ball, the type and material of the floor, and the angle and height at which the ball is dropped or thrown.

## 5. Can the "Max Force of Floor on Ball" calculation be used for other objects besides balls?

Yes, the "Max Force of Floor on Ball" calculation can be used for other objects besides balls, as long as the weight and impact force of the object on the floor are known. However, the calculation may vary depending on the shape and material of the object.

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