Projectile Motion Question!

I am having trouble with this question.

A tennis ball is thrown off the top of a building at an angle of 28 up from the horizontal, at a speed of 15m/s. If the ball hits the ground 32m from the base, how tall is the building?

Any help would be great!

Total time airborne T = X(T)/(U cos(theta)) = 32/(15*cos(28)) = 2.416149441 = 2.42 seconds.

The height of the launch point above ground h = - Uy T + 1/2 g T^2 = -15*sin(28)*2.42 + 4.9*2.42^2 = 11.65454227 = 11.7 m ANS.

NOTE, unless the ball was actually released from the roof top, this is not the height of the building. It's the height of the release point, which is more realistically 1 or 2 meters higher than the roof top.

Let us take the upward direction as positive and downward as negative. We know that the horizontal component of velocity in a projectile remains constant, which is equal to u*cosθ. Here u=15m/s and θ=28 degrees.
horizontal component=15*cos28
distance= 32m
time=distance/speed =32/(15*cos28)

now we also know S=ut + (1/2)*at^2
applying for vertical motion==>>
vertical component of velocity =15*sin28
S= -h (the minus sign because the ball falls downward and we have taken downward as negative.
a= -g (minus sign because gravity acts downwards)
-h = 15*sin28*32/(15*cos28) - (1/2)*g*(32/(15*cos28))^2

putting g as 9.8=>
-h = 17.0147018132 - 28.6051128052

=> h= 11.590410992

on x axis displacement is 32m . velocity is 15cos(28) so time of flight = distance/speed . let time be t .
on y axis take upper y as negative and lower y as positive . initial vel0city on y axis = -15sin(28)
net displacement on y axis = hieght of tower = h
h = (ut +at^2)/2 g= +10