Projectile Motion - Range of projectile

[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.

 

[voko]

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).

 

[coggo8]

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]

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

 

[voko]

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.

 

[coggo8]

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!