Finding initial projectile velocity with angle and a point in the path?

[ICanHasHelp]

Homework Statement

A ball is hit from the ground with an angle of 36 degrees above the horizontal. The ball hits a target 30m ahead and 4m above the ground. Find the initial velocity of the ball.

Homework Equations

V_avg = Δd/Δt
a_avg = Δv/Δt
Δd = v_oΔt + 1/2aΔt²

The Attempt at a Solution

Δx = 30m
Δy = 4m
V_ox = V_o * cos 36
V_oy = V_o * sin 36
Δx = v_oxΔt + 1/2a_x(Δt)²
Δx = (V_o * sin 36)Δt + 1/2a_x(Δt)²
Since a_x = 0 then
Δx = (V_o * sin 36)Δt


This question appeared on a quiz today and caught me totally by surprise. My teacher had never assigned homework questions like this ever and my textbook did not have any questions like it either. I've asked my teacher for help, but he didn't give me a clear answer. I've tried this question again at home for another two hours but I still couldn't get anything out of it. I know I have to find Δt, but I can't seem to isolate it. I would put all my attempts on the post, but I have already destroyed them in my fustration. Please help me! I'm losing my mind!

 

[w3390]

That is the correct equation for the x component of motion.

However, the reason your getting stuck is because you need another equation so that you can eliminate the unknown variable time.

This is where the y component of the motion will come in. Do this like you did to find the x component.

Remember: When the question asks for the initial velocity, it wants the velocity in the direction of launch, not a component of the velocity.

 

[ICanHasHelp]

I'm sorry I didn't initially post it, but I did find Δy, which was

Δy = VoyΔt - -4.9m/s^2

I also tried substituting Δx/Δt in place of Vox and Δy/Δt for Voy, but I still could solve for Δt because Δt was always dividing the direction.

 

[w3390]

So first of all, the equation is delta Y = (Vo_y)*t + (1/2)(-9.8)t^2.

It should not be a minus negative 4.9. You also left off the t^2.

Also, your equation for delta x should be cos(36) not sin(36).