Projectile motion finding distance

[Coffeelover]

Homework Statement

King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 12m above the moat. The rock is launched at a speed of 27m/s and an angle of 39∘ above the horizontal. How far from the castle wall does the launched rock hit the ground?

Homework Equations

sin(theta)
cos(theta)
x=1/2at^2

The Attempt at a Solution

I found the Vy and Vx components. To find time you see how long it reaches to the peak so I divided Vy/10? I found out time was t=1.69s (probably wrong) Then right there I'm stuck. I don't know what to do with the 12m either. I think I'm on the right track but maybe I'm just wrong.

 

[ruso]

Do you have an equation for y? That would be the key in incorporating the information about 12m.

 

[Coffeelover]

Do you have an equation for y? That would be the key in incorporating the information about 12m.

do you mean the y=Vy(t)-1/2at^2? so if that's the case do you plug in 12 into y? and you are solving for t?

 

[ruso]

do you mean the y=Vy(t)-1/2at^2? so if that's the case do you plug in 12 into y? and you are solving for t?

Did you get that equation from your textbook? Doesn't really look correct unless you change what the meaning of y is.

The way you have your equation, it seems like it should be Δy = Vy(t) - 1/2at^2 or y - y₀ = Vy(t) - 1/2at^2, where y is the height at time t and y₀ is your initial height.

 

[Coffeelover]

Did you get that equation from your textbook? Doesn't really look correct unless you change what the meaning of y is.

The way you have your equation, it seems like it should be Δy = Vy(t) - 1/2at^2 or y - y₀ = Vy(t) - 1/2at^2, where y is the height at time t and y₀ is your initial height.

I was talking it over with my friend and he gave that equation. I don't understand what you're trying to solve for in that equation.

 

[ruso]

I was talking it over with my friend and he gave that equation. I don't understand what you're trying to solve for in that equation.

Well, you want to know at what time t the rock the hits the ground, as it is a necessary part in finding your final answer: "How far from the castle wall does the launched rock hit the ground?"

You can't solve for t unless you have an equation for the height of the projectile (y - y0 = Vy(t) - 1/2at^2). If you know your starting height and the height at which you want to find t, you can solve for t using that equation if you keep in mind what y and y0 is in the equation.