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## 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.