Physics Projectile (zero angle) word problem? PLEASE HELP? 10 POINTS

Joe stands on a bridge kicking stones into the water below. If he kicks a stone with a horizontal velocity of 2.10 m/s, and it lands in the water a horizontal distance of 7.50 m from where Joe is standing, what is the height of the bridge? Enter m as unit.

I don't understand... I think I did everything right, followed the formulas correctly I think... the answer I got was 5.3 m, or something along those lines. Please someone show me how to do this? I promise 10 points for your time and effort!

If we assume that the horizontal speed of the stone is constant, we can find the time taken for the stone to land in the water:
speed = distance / time
2.1 = 7.5 / t
t = 7.5 / 2.1
Then assuming that the stone had no initial vertical speed, we can say that vertically, the stone fell from rest over 7.5 / 2.1 seconds.
Using S = ut + 0.5at^2
S = 62.5m

First figure out how long it takes the stone to fall. Since it continuously travels at 2.10 m/s horizontally until it lands...

distance(horizontal) = velocity(horizontal) x time (constant velocity)
d(x) = v(x) * t
solve for t...
t = d(x)/v(x)

then write the equation for the distance it fell due to acceleration
distance (vertical) = 0.5 * acceleration * square (time)
d(y) = 0.5 * at^2

substitute for t from the first equation
d(y) = 0.5 * a * [d(x)/v(x)]^2
= 0.5 * (9.81 m/s^2) * (7.50 m / 2.10 m)^2
= 62.6 m

Applying law's of motion in the horizontal direction,
s = ut + (1/2)a×t^2
7.5 = 2.1t + (1/2)×0×t^2 ,since there is no acceleration in the horizontal direction
Gives,
t = 3.57 sec

Applying law's of motion in the vertical direction,
s = 0×3.57 + (1/2)×9.8×3.57^2
=62.5m (answer)

For all physics assignment help and detailed solution visit www.tutornik.com

for horizontal motion
a(x) = 0 ,u(x) = 2.10m/s , x=7.50
x=ut+1/2at^2
7.50=2.10*t
t = 3.57s

for vertical motion
a(y) = g , u(y) = 0 , t=3.57s
y = ut+1/2at^2
y = 1/2*9.8*(3.57)^2
= 62.5m
so bridge is 62.5m high

God i hate math