Agirl standing on a bridge throws a stone vertically downward with an initial velocity of 15.0 m/s into the river below. if the stone hits the water 2.00 seconds later, what is the height of the bridge above the water?
dont give the answer, just tell me how to do it pls

Take [down] as the positive direction. 

h = vt + 1/2 g t^2 

v is initial vertical speed 
g is acceleration due to gravity 

h = 15.0 m/s (2.00 s) + 1/2 (9.80 m/s^2 )(4 s^2) 
h = 30.0 m + 19.6 m 
h = 49.6 m 

The bridge is 49.6 m above the water.

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Just divide the x axis (the top) by 2 or multiply the y axis (the bottom) by 2