# Projectile Motion - Range of projectile

## 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.

ehild
Homework Helper
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

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.

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!