Need help finding a derivation of the range formula

[Yorganda]

I need to find a derivation of the range formula for a projectile in the absence of air.

i know that the range formula is Range=Vhxt
i know that Vh=vcostheta

but in having trouble understanding how V=Vo+at can be re-arranged to = 2Vsintheta/g=t where a=g=9.8m/s/s (gravity)

i know V=Vo+at can be arranged to equal V-Vo/g=t
but this is where i get stuck, i don't know how to get to V-Vo=2Vsintheta

i know that vsintheta = vertical velocity

If anyone can help it would be appreciated heaps

 

[cellophane]

Because of the symmetry of the parabola, the ball will hit the ground with a vertical velocity of -Vo*sin(theta). So the net change in velocity is 2*Vo*sin(theta) which takes t=2*Vo*sin(theta)/g
That's a pretty hand-wavy explanation though.

 

[sir-pinski]

The range formula for a projectile in the absence of air is derived from the equations of motion and the principles of projectile motion. It is given by:

Range = Vx * t

Where Vx is the horizontal component of the initial velocity and t is the time of flight.

To understand how this formula is derived, we need to break it down into its components. The horizontal motion of a projectile is constant, which means that the horizontal velocity remains the same throughout the motion. This is given by:

Vx = Vo * cos(theta)

Where Vo is the initial velocity and theta is the angle of projection.

Now, let's consider the vertical motion of the projectile. The vertical velocity changes due to the acceleration of gravity. We can use the equation of motion for vertical motion to find the time of flight, t:

V = Vo + at

Where V is the final vertical velocity, Vo is the initial vertical velocity, a is the acceleration due to gravity (which is -9.8 m/s^2) and t is the time of flight.

Since the projectile starts and ends at the same height, the vertical displacement is zero. Using the equation of motion for vertical motion, we can find the time of flight, t:

0 = Vo * t + (1/2) * a * t^2

Solving for t, we get:

t = 2 * Vo * sin(theta) / a

Substituting this value of t in the range formula, we get:

Range = Vx * t

= (Vo * cos(theta)) * (2 * Vo * sin(theta) / a)

= (2 * Vo^2 * sin(theta) * cos(theta)) / a

= (Vo^2 * sin(2 * theta)) / a

This is the final form of the range formula, which is derived from the equations of motion and the principles of projectile motion. I hope this helps in understanding the derivation of the range formula.