# How to get R= 2 ( square root ) h1h2

• DLeuPel
In summary: I can’t. Just looked at the exercise again. One of the follow up questions is “ explain why you didn’t need to know...
DLeuPel
Member advised to use the formatting template for all homework help requests
Problem :

A ball is let down a ramp on top of a table with initial velocity of 0 ms-1. When it reaches the end of the ramp, it is launched horizontally. Knowing that we don’t take air resistance or friction into account, and that the height of the ramp is h1, and that of the table is h2 relative to ground level,how do we get to the equation R (range or distance traveled from the end of the ramp) = 2√(h1h2)

Equations needed:

Since it is a projectile motion related exercise, the equations needed are Vv= vcosΘ , Vh= vsinΘ
and all the cinematic equations.

My approach

Since it’s projected horizontally i tried solving it taking into account that Vv = 0 and that Vh= t1h1. So I had to multiply gt1 by time (t2) to obtain the range. And t is equal to v/a or an alternative, t= √(2h2/g). But I wasn’t able to get to the equation R= 2√(h1h2). How do I ?

Last edited:
DLeuPel said:
A ball is let down a ramp on top of a table with initial velocity of 0 ms-1. When it reaches the end of the ramp, it is launched horizontally. Knowing that we don’t take air resistance or friction into account, and that the height of the ramp is h1, and that of the table is h2 relative to ground level,how do we get to the equation R (range or distance traveled from the end of the ramp) = 2√(h1h2)

Since it is a projectile motion related exercise, the equations needed are Vv= vcosΘ , Vh= vsinΘ
and all the cinematic equations. Since it’s projected horizontally i tried solving it taking into account that Vv = 0 and that Vh= gh1. So I had to multiply gh1 by time (t) to obtain the range. And t is equal to v/a or an alternative, t= √(2h2/g). But I wasn’t able to get to the equation R= 2√(h1h2). How do I ?

How did you get ##v_h = gh_1##?

PeroK said:
How did you get ##v_h = gh_1##?
I made I mistake, it is meant to be gt, i’ll correct it.

DLeuPel said:
I made I mistake, it is meant to be gt, i’ll correct it.

For the motion down the ramp, do the kinematic equations help? Or, is there another way to study this motion?

PeroK said:
For the motion down the ramp, do the kinematic equations help? Or, is there another way to study this motion?
I guess you could use the kinetic energy, but I used the kinematic equations so I use the equations that I already seen in the book, despite the fact that I can use them.

DLeuPel said:
I guess you could use the kinetic energy, but I used the kinematic equations so I use the equations that I already seen in the book, despite the fact that I can use them.

So, what do you have for the motion down the ramp?

PeroK said:
So, what do you have for the motion down the ramp?
since v = u + at and u = o I get that v = at or v = gt. V would be the initial velocity for the projectile motion. I can use this equation since the exercise supposed that there is no friction it is approximately like if it would be free falling. Any thoughts ?

DLeuPel said:
since v = u + at and u = o I get that v = at or v = gt. V would be the initial velocity for the projectile motion. I can use this equation since the exercise supposed that there is no friction it is approximately like if it would be free falling. Any thoughts ?

How are you going to find ##t##?

Note, also, you are told nothing about the shape or angle of the ramp.

PeroK said:
How are you going to find ##t##?

Note, also, you are told nothing about the shape or angle of the ramp.
What the exercise wants me to do is not solve the problem, which would mean knowing the maximum range that the ball will travel. What the exercise wants me to do is to demonstrate that the problem can be solved by using R = 2√(h1h2).

DLeuPel said:
What the exercise wants me to do is not solve the problem, which would mean knowing the maximum range that the ball will travel. What the exercise wants me to do is to demonstrate that the problem can be solved by using R = 2√(h1h2).

What do you know about energy?

PeroK said:
What do you know about energy?
The basics. KE= mv2/2 and PE= mgh. And of course ME= KE + PE

DLeuPel said:
The basics. KE= mv2/2 and PE= mgh. And of course ME= KE + PE

Okay. Use that for the motion down the ramp.

PeroK said:
Okay. Use that for the motion down the ramp.
I can’t. Just looked at the exercise again. One of the follow up questions is “ explain why you didn’t need to know the mass of the ball for your answer”.

DLeuPel said:
I can’t. Just looked at the exercise again. One of the follow up questions is “ explain why you didn’t need to know the mass of the ball for your answer”.

What is this word "can't"? At least try.

PeroK said:
What is this word "can't"? At least try.
I got the equation right, thank you. Also realized how to accomplish the same thing with kinematics. Thanks for the help

## 1. How do I calculate R= 2 (square root) h1h2?

To calculate R= 2 (square root) h1h2, you will need to use the formula R= 2√h1h2. This means that you need to take the square root of the product of h1 and h2, and then multiply it by 2.

## 2. Can I use a calculator to find the value of R= 2 (square root) h1h2?

Yes, you can use a calculator to find the value of R= 2 (square root) h1h2. Simply enter the values for h1 and h2 into the calculator, take the square root of the product, and then multiply it by 2 to get the value of R.

## 3. What units should I use for h1 and h2 when calculating R= 2 (square root) h1h2?

When using the formula R= 2 (square root) h1h2, it is important to use consistent units for h1 and h2. Make sure to either use both values in meters or both in centimeters, for example, to get an accurate result for R.

## 4. How can I use R= 2 (square root) h1h2 in my scientific research?

R= 2 (square root) h1h2 can be used in various scientific research areas, such as geometry, physics, and engineering. It is commonly used to calculate the radius of a cylinder or cone, and can also be used in fluid dynamics and surface area calculations.

## 5. Is there a simpler way to represent R= 2 (square root) h1h2?

Yes, there is a simpler way to represent R= 2 (square root) h1h2. It can also be written as R= 2√(h1 x h2) or R= 2√(h1h2), depending on the notation preference. Both forms mean the same thing and can be used interchangeably.

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