Cannon ball kinematics proof

[msadegian]

A cannon ball is fired from an angle o with an intial velocity of v. The hill sklopes down with an angle of O. Prove that the horizontal distance the cannon ball travels is given by dx= 2v²cos(o)sin(O+o)/gcos(O)



2. Equations
sin(O+o)=sin(O)cos(o)+sin(o)cos(O)
d=1/2gt²

The Attempt at a Solution

Please help me! Hopefully the picture helps!

 

[LowlyPion]

Welcome to PF.

You might want to consider the point at which it hits as the intersection of the function that describes the projectile's flight with the equation that describes the slope of the incline.

For instance for 45 degrees you know that dx = dy.

Or more generally for angle θ you have Dy = tanθ *dx

 

[nlightner]

I would like to thank you for providing the necessary information and equations to prove the horizontal distance of a cannon ball fired from an angle o with an initial velocity of v on a hill that slopes down with an angle O. Let's break down the problem step by step.

First, we need to understand the motion of the cannon ball. From the given information, we can assume that the cannon ball is being fired in a projectile motion, with an initial velocity v at an angle o. This means that the horizontal component of the velocity is vcos(o) and the vertical component is vsin(o).

Next, we need to consider the slope of the hill, which is given by the angle O. This means that the slope of the hill is perpendicular to the ground, and therefore, the vertical component of the velocity will change due to the gravitational force acting on the cannon ball. However, the horizontal component of the velocity remains constant throughout the motion.

Now, let's consider the horizontal distance traveled by the cannon ball. We can use the equation d=1/2gt^2 to calculate this distance, where d is the horizontal distance, g is the acceleration due to gravity, and t is the time. However, we need to find the value of t in terms of v, o, and O.

To do this, we can use the equation of motion for the vertical component of the velocity, which is v=gt. Solving for t, we get t=v/g. Now, we can substitute this value of t in the equation for horizontal distance, and we get d=1/2g(v/g)^2 = v^2/2g.

But, this is only the horizontal distance traveled in a straight line. We need to take into account the angle of the hill, O, and the angle at which the cannon ball is fired, o. To do this, we can use the trigonometric identity sin(O+o)=sin(O)cos(o)+sin(o)cos(O).

Substituting this in our equation for horizontal distance, we get d= v^2/2g * sin(O)cos(o)+sin(o)cos(O). Simplifying this further, we get d= v^2/2g * sin(o+O).

Finally, we need to consider the fact that the horizontal distance traveled is also affected by the horizontal component of the velocity, which is vcos(o).