# Deriving the expression for average vertical force

• jisbon
In summary: Re your first question: why isn't that obvious?Can't you deduce ##v_y## from high-level information about the problem?
jisbon
Homework Statement
Ball (mass m) launched horizontally at height H from ground.
Horizontal displacement at first bounce is 4/3 H and maximum height after the first bounce is 9/16 H.
Duration of contact between ball and ground is t
Relevant Equations
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Here's what I have done:

Let t1 be the time taken for the ball to reach the ground after the first bounce, and ux be the initial speed in the horizontal direction of ball.
Also, let v be the speed after the bounce.
For horizontal,
$$\dfrac {4}{3}H=U_{x}t_{1}$$
For vertical,
$$H=\dfrac {1}{2}g\left( t_{1}\right) ^{2}$$
Solving these 2, I find:
$$U_{x}=\dfrac {2}{3}gt_{1}$$
Im supposed to find the vertical average force. In this case will it be:
$$F_{avg}=\dfrac {m\left( v_{y}-u_{y}\right) }{t}$$
?
Not sure what the question is requiring me to do

What vertical force are you trying to find? gravity, air resistance, contact force with the ground?

PeroK said:
What vertical force are you trying to find? gravity, air resistance, contact force with the ground?
I will assume contact force. The question simply stated to derive the expression for vertical average force acting on ball

jisbon said:
I will assume contact force. The question simply stated to derive the expression for vertical average force acting on ball

If you don't know what force you are supposed to be calculating then I'd move on.

I'd assume it's the contact force.

Force is rate of change of momentum, I believe.

PeroK said:
If you don't know what force you are supposed to be calculating then I'd move on.

I'd assume it's the contact force.

Force is rate of change of momentum, I believe.
So this is correct I assume?
$$F_{avg}=\dfrac {m\left( v_{y}-u_{y}\right) }{t}$$

jisbon said:
So this is correct I assume?
$$F_{avg}=\dfrac {m\left( v_{y}-u_{y}\right) }{t}$$
Looks good. Although, you might need to figure out what they want you to do about gravity.

Is this the contact force or the net force?

PeroK said:
Looks good. Although, you might need to figure out what they want you to do about gravity.

Is this the contact force or the net force?
The answer provided to me was :
$$\dfrac {7m}{4t}\sqrt {2gH}$$
What I have here is:
$$F_{avg}=\dfrac {m\left( v_{y}-u_{y}\right) }{t}$$
Where uy is 0
So I found:
\begin{aligned}v_{y}=gt_{1}=g\sqrt {2H}\\ F=\dfrac {m\left( v_{y}-0\right) }{t}=\dfrac {mg\sqrt {2H}}{t}\end{aligned}
What's the error here? Thanks

Why
jisbon said:
The answer provided to me was :
$$\dfrac {7m}{4t}\sqrt {2gH}$$
What I have here is:
$$F_{avg}=\dfrac {m\left( v_{y}-u_{y}\right) }{t}$$
Where uy is 0
So I found:
\begin{aligned}v_{y}=gt_{1}=g\sqrt {2H}\\ F=\dfrac {m\left( v_{y}-0\right) }{t}=\dfrac {mg\sqrt {2H}}{t}\end{aligned}
What's the error here? Thanks
Why do you have ##u_y = 0##?

PeroK said:
Why

Why do you have ##u_y = 0##?

Because the ball is launched horizontally, so
$$u_y = 0$$

jisbon said:
Because the ball is launched horizontally, so
$$u_y = 0$$
I was hoping you wouldn't say that!

So why not take ##v_y =0##, as the ball reaches the top of the first bounce? That's where the problem statement ends.

Or, should these velocities relate to the phase of the motion you are interested in?

PeroK said:
I was hoping you wouldn't say that!

So why not take ##v_y =0##, as the ball reaches the top of the first bounce?

Or, should these velocities relate to the phase of the motion you are interested in?
Not sure what you meant.. But wasn't it constrained to the first phase when I initially subbed $$\dfrac {4}{3}H=U_{x}t_{1}$$?

jisbon said:
Not sure what you meant.. But wasn't it constrained to the first phase when I initially subbed $$\dfrac {4}{3}H=U_{x}t_{1}$$?
I'm not sure what you mean by that. If you want the force over a time interval ##t##, then you need the velocities at the start and end of that interval.

If you include an earlier phase of motion, then you need to include the time for that phase too. In this case ##t_1## would be relevant.

PeroK said:
I'm not sure what you mean by that. If you want the force over a time interval ##t##, then you need the velocities at the start and end of that interval.

If you include an earlier phase of motion, then you need to include the time for that phase too. In this case ##t_1## would be relevant.
So from what I understand, I need to find the velocity JUST before the bounce and JUST after the bounce?
So JUST before the bounce, velocity of y = $$g\sqrt {2H}$$
How do I find the velocity JUST after the bounce in this case?

jisbon said:
So from what I understand, I need to find the velocity JUST before the bounce and JUST after the bounce?
So JUST before the bounce, velocity of y = $$g\sqrt {2H}$$
How do I find the velocity JUST after the bounce in this case?
Re your first question: why isn't that obvious?

Can't you deduce ##v_y## from high it bounces?

PeroK said:
Re your first question: why isn't that obvious?

Can't you deduce ##v_y## from high it bounces?
Sorry, wasn't thinking clearly.
So can I say that \begin{aligned}0=v^{2}_{y}+2\left( -g\right) \left( \dfrac {9}{16}H\right) \\ v^{2}_{y}=\dfrac {18}{16}gH\end{aligned}

PeroK
Yes. Be careful with signs now.

Ok.. So force is derived to be:

$$\dfrac {m\left( \sqrt {\dfrac {18}{16}gH}-g\sqrt {2+t}\right) }{t}$$

How does one simplify to the suggested answer? Can't get my head around it

jisbon said:
Ok.. So force is derived to be:

$$\dfrac {m\left( \sqrt {\dfrac {18}{16}gH}-g\sqrt {2+t}\right) }{t}$$

How does one simplify to the suggested answer? Can't get my head around it
I can't see where you got that second term from. The speed as it leaves the surface can be determined by the same method you got the incident velocity.

In general, it is always ##v^2 = 2gh##. It's only ##h## that changes each time.

jisbon
PeroK said:
I can't see where you got that second term from. The speed as it leaves the surface can be determined by the same method you got the incident velocity.

In general, it is always ##v^2 = 2gh##. It's only ##h## that changes each time.
Made a mistake in Latex from my post, I meant 2gH.
Also, I managed to get the solution after realising that it should be negative in the second part since it is bouncing in a different direction.
Thanks for the help.

## What is average vertical force?

Average vertical force is the force exerted in the vertical direction over a certain period of time, which is calculated by dividing the total vertical force by the time interval.

## Why is it important to derive the expression for average vertical force?

Deriving the expression for average vertical force allows us to understand the relationship between force, mass, and acceleration in the vertical direction. This can be applied to various real-world situations, such as calculating the lift force on an airplane or the gravitational force on an object.

## How is the expression for average vertical force derived?

The expression for average vertical force is derived using Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration. By considering the forces acting on an object in the vertical direction, we can derive the expression for average vertical force.

## What are the variables involved in the expression for average vertical force?

The variables involved in the expression for average vertical force are mass, acceleration, and time. Mass is the amount of matter in an object, acceleration is the rate of change of velocity, and time is the duration over which the force is applied.

## Can the expression for average vertical force be applied to any object?

Yes, the expression for average vertical force is a fundamental equation in physics and can be applied to any object, regardless of its shape, size, or composition.

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