# Projectile motion with differing heights

Hedgehog1

## Homework Statement

A cannon is fired at a target on a cliff that is 400 meters away. The cliff has an elevation of 150 meters relative to the cannon. The initial velocity of the cannonball is 100 m/s. Ignoring wind resistance calculate the two possible angles and their respective times.
X= 400m
Y=150m
V_naught=100m/s
A= -9.81 m/s^2

## Homework Equations

Time X = Time Y

X=V_naught*cos(theta)*t
Y= V_naught*sin(theta)*t - 0.5*a(t)^2

1/(cos(theta)0^2=1+ (tan(theta))^2

## The Attempt at a Solution

t= 400/100cos(theta)

0= -0.5*a(t)^2 + V_naught*sin(theta)*t -150

0= -0.5*a(400/100cos(theta))^2 + V_naught*sin(theta)*400/100cos(theta) -150

I get stuck here. I'm not really quite sure what to do next. I'm pretty sure I have to use the quadratic equation, but don't know how.

Homework Helper
Hi Hedgehog, welcome to PF.

Substitute vnought=100 into your equation and simplify. You also know that a=g, the gravitational acceleration.

ehild

Hedgehog1

I substituted all my v0=100 and simplified to get this:

-0.5*-9.81(16/1+(tan(theta))^2) + 400*tan(theta) -150=0

Edit:
I'm not sure entirely how I would simplify further.

Homework Helper
Your equation Y= V0*sin(theta)*t - 0.5*a(t)^2 is a bit incorrect, it has to be Y= V0*sin(theta)*t + 0.5*a(t)^2, with a=-9.81 m/s^2.
Using the parentheses correctly, you get
-0.5*9.81*16(1/cos2(θ))+ 400*tan(θ) -150=0.
You wrote among the relevant equations that 1/cos2(θ)=1+tan2(θ). Use it, substitute 1+tan2(θ)for 1/cos2(θ). You get a quadratic equation for tan(θ).

ehild

Hedgehog1
Thank you for your help! I understand what to do now.

Homework Helper
Thank you for your help! I understand what to do now.

You are welcome

ehild