Theoretical Range Equation Derivation

[ccsmarty]

Homework Statement

Derive an expression for calculating the theoretical range of the ball in terms of y, g, theta, and v0.

Homework Equations

The Attempt at a Solution

I've tried just about everything to figure this out. I started with the motion in the x-direction, and plugged the second-to-last equation in for v0x in the second equation, and then solved for "t". Then I plugged the "t" into the second equation (but for the y-direction), and tried solving for "x". But I kept getting really weird numbers. Can someone please help me out?

 

[mgb_phys]

Motion in the x direction is at a constant speed, so what you are looking for is the time given by motion in the y direction.
Use s = ut + 1/2 at^2 and v^2 = u^2 + 2as to get an equation for time as a function of initial vertical velocity component. Remember that final (vertical) velocity is zero at the top of the curve.
Then plug this into the horizontal component of velocity to get a range.

 

[m2003]

I understand your struggle with this problem. It can be frustrating when equations don't seem to work out. However, it's important to remember that theoretical equations are just models and may not always give accurate results in real-world situations.

In this case, the theoretical range equation is derived from the equations of motion for a projectile. It takes into account the initial velocity (v0), the angle of launch (theta), and the acceleration due to gravity (g). The equation is as follows:

R = (v0^2 * sin(2*theta)) / g

Where R is the range of the projectile.

To derive this equation, you can start by breaking down the motion of the projectile into its x and y components. In the x-direction, the projectile will have a constant velocity (v0x) and no acceleration. This can be represented by the equation:

x = v0x * t

Where x is the distance traveled in the x-direction and t is the time.

Next, in the y-direction, the projectile will have a constant acceleration due to gravity (g) and an initial velocity (v0y). This can be represented by the equation:

y = v0y * t + (1/2) * g * t^2

Where y is the distance traveled in the y-direction.

Now, we can use the relationship between time and distance to eliminate t from these equations. This relationship is:

t = x / v0x = y / v0y

By substituting this into the equations for x and y, we can get the following equations:

x = (v0^2 * sin(theta) * cos(theta)) / g

y = (v0^2 * sin^2(theta)) / (2 * g)

Finally, to find the range (R), we need to find the value of x when y is equal to zero (since the projectile will hit the ground at that point). This can be done by setting the y-equation equal to zero and solving for x:

0 = (v0^2 * sin^2(theta)) / (2 * g)

x = (v0^2 * sin(theta) * cos(theta)) / g

By substituting x and y into the equation for range (R), we get the theoretical range equation:

R = (v0^2 * sin(2*theta)) / g

I hope this helps you understand