# Projectile Motion cannon off a cliff

• Towk667
In summary, you are given the distance from the bottom of the cliff, the initial velocity, and the height of the cliff. You use an equation to find the angle at which the object was shot.

#### Towk667

Okay there is an object being shot from a cannon at an angle off a cliff. You are given the distance from the bottom of the cliff, the initial velocity, and the height of the cliff. What is an equation that would use these knowns to give you the measure of the angle that the object was shot at. Can someone help me with this?

think about how you would find the range of a projectile if you were given the height of the cliff and the initial velocity.

you can then work backwards to find theta.

Towk667 said:
Okay there is an object being shot from a cannon at an angle off a cliff. You are given the distance from the bottom of the cliff, the initial velocity, and the height of the cliff. What is an equation that would use these knowns to give you the measure of the angle that the object was shot at. Can someone help me with this?

These equations might be a useful resource for you:
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

Okay, I get an equation with both sine and cosine terms that I can't get rid of. Can someone just show me how this is done?

Towk667 said:
Okay, I get an equation with both sine and cosine terms that I can't get rid of. Can someone just show me how this is done?

Typically the trig identity

2*sinθ*cosθ = sin2θ

is of use in resolving Range equations.

Without seeing your work, there's no useful way to help you with what you are doing that I can see.

show us the equation and we'll help...

or just remember that sin(x)/cos(x) = tan(x)

edit: yeah or the trig identity posted above

Using earlofwessex's idea, I start with R=((V0/g)(V0sinΘ+sqrt((V0sinΘ)^{2}+2gh) where V0 is initial velocity, Θ is the angle, R is range, g is acc due to gravity, and h is the height of the cliff.

I tried to solve this equation for Θ and get:

(gR)/(V0^{2})=.5sin2Θ+sqrt(sin^{2}Θ+((2gh)/V0))

Am I using the wrong equation or there another trig identity or what?

Towk667 said that the distance from the bottom of the cliff and the height of the cliff were given seperately. So that must mean the cannon shot the ball from a distance from the end of the cliff. Keeping this in mind, I found the horizontal range, i.e., the distance from the cannon(considering its distance from the end of the cliff) to the place where the ball hits the ground:

{[ucosӨ * sqrt(u^2sin^2Ө + 2gh)] +[2u^2sin2Ө]}/2g = R

I don't know what to do ahead of this, but I'm quite sure this is the correct equation.

I've got another question related to projectile motion:
"What is the average velocity of a projectile between the instants it crosses half the maximum height? It is projected with a speed 'u' at an angle Ө with the horizontal"

(a) u sinӨ (b) u cosӨ (c) u tanӨ (d) u

Could I get some help on this one? I don't even know where to start.

modulus said:
I've got another question related to projectile motion:
"What is the average velocity of a projectile between the instants it crosses half the maximum height? It is projected with a speed 'u' at an angle Ө with the horizontal"

(a) u sinӨ (b) u cosӨ (c) u tanӨ (d) u

Could I get some help on this one? I don't even know where to start.

I think I can help with this one. Think about the formula to find maximum height and then use that knowledge and a formula for final velocity that doesn't use time to get your answer.

Towk667 said:
I think I can help with this one. Think about the formula to find maximum height and then use that knowledge and a formula for final velocity that doesn't use time to get your answer.

I did what you said and got the following equation for the final velocity:
sqrt{ [ (2g (u sinӨ)^2) + (2g (u cosӨ)^2) - ((u sinӨ)^2)] / 2g }

Now, what do I do? I don't think I can use the final velocity to evaluete the average velocity, right?
Is there any other method in which I can find the total distance and total time?

## 1. What is projectile motion?

Projectile motion is the motion of an object through the air under the influence of gravity. It follows a parabolic path due to the acceleration of gravity.

## 2. How does a cannon off a cliff demonstrate projectile motion?

A cannon off a cliff demonstrates projectile motion because the cannonball is launched at an angle from the cliff and falls under the force of gravity, following a parabolic path.

## 3. What factors affect the trajectory of a projectile off a cliff?

The factors that affect the trajectory of a projectile off a cliff include the initial velocity, angle of launch, and the acceleration of gravity. Air resistance can also have an impact on the trajectory.

## 4. How can you calculate the range of a projectile off a cliff?

The range of a projectile off a cliff can be calculated using the equation: range = (initial velocity squared * sine of twice the launch angle) divided by the acceleration of gravity.

## 5. Can the projectile's trajectory be affected by the height of the cliff?

Yes, the height of the cliff can affect the projectile's trajectory. A higher cliff will result in a longer flight time, which can impact the range and landing position of the projectile.