H(t)=15-10t-16t^2 if a snowboarders horizontal velocity is 10feet per second, how far from the base of the overhang will she land? 15 equals initial height of overhang, -10 is the initial vertical velocity and t is the time.

H ( t ) = 15 - 10 t - 16 t² = - 16 t² - 10 t + 15
15 ft is the initial height, or H ( 0 ) and - 10 ft / s  is the initial vertical velocity and 10ft/s is horizontal velocity. As for the time:
t 1/2 = (- b +/- √( b² - 4 a c ) )/ 2 a
t 1/2 = ( 10+/-√(100 + 960 ) ) / ( -32 ) =
= ( 10 - 32.56 ) / ( -32 ) = - 22.56 / ( - 32 ) = 0.705 s  ( another solution is negative )
d = vo x · t = 10 ft/s · 0.705 s = 7.05 ft.
She will land 7.05 ft from the base of the overhang. 

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