1. The problem statement, all variables and given/known data A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 4.0s later. how high is the cliff 2. Relevant equations .5mv^2+mgy = E v = v0 + at v = d/t 3. The attempt at a solution The question says after the rock is DROPPED so i need to figure out the time it took to drop and the time it took for the sound to travel back and those two times added together will be 4 sec t1 is the time it takes for the rock to hit the water. using energy mgh = .5mv^2 gh = .5v^2 to find v v = v0 + at v = gt so gh = .5g^2t^2 2h=gt^2 2h/g = t^2 t1= sqrt (2h/g) to find t2 v = d/t v=h/t t2= h/v where v is speed of sound t1 + t2 = 4 sqrt(2h/g) + h/v = 4 sqrt(2h/g) = 4 - h/v 2h/g = (4-h/v)^2 2h/g = 16 - 8h/v + h^2/v^2 2h/g +8h/v = 16 - h^2/v^2 2hv/gv + 8hg/gv = 16 - h^2/v^2 [(2hv+8hg)/gv] = 16 - h^2/v^2 [(2v+8g)/gv]h = 16 - h^2/v^2 h^2/v^2 + [(2v+8g)/gv]h - 16 = 0 (1/v^2)h^2 +[(2v+8g)/gv]h -16 = 0 (1/(343^2))h^2 + [(2*343+8*9.8)/(9.8*343)]h -16 = 0 (8.50x10^-6)h^2 + .227 h -16 = 0 using quadratic formula [-.227 +/- sqrt((-.227^2)-4(8.50x10^-6)(-16))]/(2* (8.50x10^-6) (-.227 +/- .228)/ (1.7*10^-5) -.227 + .228/ (1.7*10^-5) = 58.8m = h mastering physics wants it to 2 sig figs so i submitted 59 m but it was marked wrong. where did i make a mistake?