Aball is thrown horizontally at a speed of 24 meters per second from the top of the building. if the ball hits the ground 4.0 seconds later approximately how high is the building?

The problem indicates that the ball was thrown horizontally. Therefore Vox is 24 m/s and Voy = 0
Yo = initial eight where the ball is coming from
Y is the final height or the ground

The formula will be like this, 
Y=Yo+Voy t- (1/2) gt^2 with that said Voy t = 0 since Voy=0,
With the height or Y = 0

0=Yo-(1/2)gt^2
Yo= (1/2)gt^2

then solve the equation and it will look like this,

Yo= (1/2)(9.8 m/s^2)[(4s)^2]

Then the answer would be

Yo = 78.4 m