# Projectile Motion - Range of projectile

• coggo8
In summary, the problem consists of two parts: finding the distance a projectile fired from a cannon with a velocity of 120 ms^-1 will travel from the top of a cliff, and finding the distance the projectile will travel when the cannon is aimed at a 45° angle. Using the equations Δy = u(y)*t + 1/2*a(y)*t^2 and Δx = u(x)*t, the solution for Part 1 is found to be 663.9 m. For Part 2, the equation -150 = 84.85*t + 1/2(-9.8)*t^2 is used to find the time and then the distance is calculated using Δx=u(x)*
coggo8

## Homework Statement

My problem consists of two parts, the first being how far from the top of a cliff will a projectile (fired from a cannon) land with a velocity of 120 ms^-1, fired horizontally from cliff height of 150m.
The second part is where I am struggling, the cannon is now aimed 45° above horizontal, how far will the projectile travel now?

Known data for part 1.
Δy = -150
u(y) = 0 ms^-1
a(y) = -9.8 ms^-2
u(x) = 120 ms^-1

## Homework Equations

Δy = u(y)*t + 1/2*a(y)*t^2
Δx = u(x)*t

x any y component equations.
u(x) = u*cos(θ)
u(y) = u*sin(θ)

## The Attempt at a Solution

Part 1 I found quite simple so hopefully this is correct..

Δy = u(y)*t + 1/2*a(y)*t^2
-150 = 0*t + 1/2(-9.8)*t^2 (substitute above data)
-150 = -4.9*t^2
t^2 = -150/-4.9
t = 5.53 seconds

Δx = u(x)*t
Δx = 120*5.53
Δx = 663.9 m

Part 2.

u(x) = 120*cos(45)
u(x) = 84.85 ms^-1 (2 d.p.)

u(y) = 120*sin(45)
u(y) = 84.85 ms^-1

This is as far I got (finding the x and y components of the initial velocity).
My issue is that I am not sure how to account for the fact that the cannon is elevated 150 m along with the angle change from horizontal to 45°.
Any help will be greatly appreciated.

The equation you used in Part 1 for ##\Delta y## works in Part 2, too. Except that ##u_y## is not zero in Part 2 (as you have already found).

Which of these should the equation look like?
-150 = 84.85*t + 1/2(-9.8)*t^2
or
Δy = 84.85*5.53 + 1/2(-9.8)*(5.53)^2

With the first I have an issue because I am unable to find a value for t and my value for Δy (-150) will also be incorrect as the projectile is now being fired at an angle.
With the second I am using an incorrect value for t as this was the value used for the horizontally fired projectile.

The cannon is on the top of the hill at hight of 150 m, and its tube is directed at 45°angle. The velocity of the cannon ball has a positive vertical component so it takes longer time to reach the ground as previously, when it was projected horizontally. Determine the time from the equation -150 = 84.85*t + 1/2(-9.8)*t^2. As the cannon is small with respect to the hill you can take Δy=-150 m again.

ehild

coggo8 said:
Which of these should the equation look like?
-150 = 84.85*t + 1/2(-9.8)*t^2
or
Δy = 84.85*5.53 + 1/2(-9.8)*(5.53)^2

With the first I have an issue because I am unable to find a value for t and my value for Δy (-150) will also be incorrect as the projectile is now being fired at an angle.

Firing at the angle does not change the vertical displacement. Why are you unable to find a value for t? It is a quadratic equation.

voko said:
Firing at the angle does not change the vertical displacement. Why are you unable to find a value for t? It is a quadratic equation.

Of course, I hadn't thought of quadratics (I am just learning kinematics via correspondence). Thank you for your help!

## 1. What is projectile motion?

Projectile motion is the motion of an object through the air that is influenced by gravity. It follows a curved path known as a parabola.

## 2. What factors affect the range of a projectile?

The range of a projectile is affected by its initial velocity, the angle at which it is launched, and the force of gravity. Air resistance can also play a role.

## 3. How do you calculate the range of a projectile?

The range of a projectile can be calculated using the formula R = (v^2 sin2θ)/g, where R is the range, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## 4. How does the range of a projectile change with launch angle?

The range of a projectile is the maximum when it is launched at a 45 degree angle. As the launch angle increases or decreases from 45 degrees, the range decreases.

## 5. Can the range of a projectile be affected by air resistance?

Yes, air resistance can affect the range of a projectile. As air resistance increases, the range decreases due to the opposing force acting on the projectile.

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