Projectile motion (time cut in half)

[reshmaji]

Homework Statement

Two buildings A & B are each of height H as measured from the ground & are located a distance L apart. A tennis ball is shot horizontally with a velocity v_o from building A such that it just hits the bottom of building B before hitting the ground. If building B is moved to half the distance (1/2)(L) & the ball is launched with the same horizontal velocity, where on building B does it hit the side, measured from the ground?

Homework Equations

I think they are:
x-x_o = (v_ocos(theta))t
y-y_o = (v_osin(theta))t - 1/2(g)t²

The Attempt at a Solution

Where the ball would be at L/2 = x/2 & results in t/2 because x is proportional to time in the equation.
So apply half the time to the y-axis equation, but since it's 2 terms with different degrees of t (one is t¹, the other t²) I'm not sure what I can assert about how this effects y.

 

[reshmaji]

I just realized that the y-axis won't use the equation I posted, the y-axis is a one dimensional motion... I'll try this again

 

[reshmaji]

Okay, is this correct:
We can deduce that it'll be half the time from my reasoning in the first post.
the y-axis equation we need to use is: Δy = v_ot + ½at²
since v_o = 0, Δy = ½at²,
if we apply t/2 we get Δy = ½a(½t)² =(¼)½at², so we'll end up with ¼*Δy, giving us 3/4H from the ground.

Is this all sound reasoning?

 

[cnh1995]

Okay, is this correct:
We can deduce that it'll be half the time from my reasoning in the first post.
the y-axis equation we need to use is: Δy = v_ot + ½at²
since v_o = 0, Δy = ½at²,
if we apply t/2 we get Δy = ½a(½t)² =(¼)½at², so we'll end up with ¼*Δy, giving us 3/4H from the ground.

Is this all sound reasoning?

Looks good!