physics
posted by Anonymous .
a ball is thrown horizontally from a hill 29.0m high at a velocity of 4.00m/s. find the distance between the base of the hill and the pointwhere the ball hits the ground.

intial Vy=0 (since it's thrown horizontal) change in y = 29.0m Ay=9.81 m/s^2
find the time used by equate :
change in Y= intial Vy times time +1/2 Ay times time squared so 29.0=1/2(9.81)t^2
t^2 = .91s so t=2.43s
since time used for y is same as x
change in x = 1/2 (intial Vx+ final Vx) times time
in progectile motion ini Vx = fin Vx
so change in x = time(2.43s) times intial vx (4.00m/s))