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.

 

[ehild]

Hi Hedgehog, welcome to PF.

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

ehild

 

[Hedgehog1]

Thanks for the quick reply!

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.

 

[ehild]

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.

 

[ehild]

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

You are welcome

ehild