Projectile Motion Problem: Finding Object Speed for Given Height and Angle

[AlexT89]

Homework Statement

This is a projectile motion problem, providing only the maximum height the object reaches and the angle that it leaves the ground, and asking to find the speed at which the object must leave in order to reach that max height:

The object leaves the ground at 45 degrees and reaches a max height of 12 feet. With what speed must the object leave the ground to reach that height?

Homework Equations

y-direction:
where V_oy = V_o sin 45,
y displacement (12 feet) = V_oyt + (0.5)(-9.8m/s₂)t²

V_o = (V_oy) / (sin 45)

The Attempt at a Solution

I feel like I'm going in circles every time I attempt to solve this problem, almost like I don't have enough information, even though I know that I do.

I've gotten as far as this:

3.66 m = (V_o)(sin 45)t - 4.9t²

I will appreciate any help you can give me. I think I am looking at it wrong, but I can't figure out how else to do so.

Thanks in advance,
Alex

 

[method_man]

When projectile reaches it's highest point, shouldn't the v_y be equal to zero?

 

[AlexT89]

When projectile reaches it's highest point, shouldn't the v_y be equal to zero?

Right, so if I used the equation:

V_y² = V_oy² + 2(-9.8)(y-disp)

it would allow me to solve for V_oy, and by extension V_o.
Is that correct?

 

[AlexT89]

That worked out, I got the answer that it gives in the back of the book. I knew I was overlooking something. Thanks for your help, method man.

 

[method_man]

Right, so if I used the equation:

V_y² = V_oy² + 2(-9.8)(y-disp)

it would allow me to solve for V_oy, and by extension V_o.
Is that correct?

Yes something like that. Yours y-disp is maximum height and of course V_y=0.

You could also do it like this:
v_y=v_0y-g*t
And when height is maximum, v_y=0 so you get v_0y=g*t. From here, you can extract either t or v_0y and put it in that first equation: y = V_oyt + (0.5)(-9.8m/s2)t².