Arock is thrown upward with a velocity of 21 meters per second from the top of a 43 meter high cliff, and it misses the cliff on the way back down. when will the rock be 5 meters from the water, below? round your answer to two decimal places.

5.66 seconds from when the rock was thrown. First construct an expression to represent the distance from the water to the rock. First account for the starting distance. d = 43 + ? Now add in the distance based upon the initial velocity d = 43 + 21t + ? Now add in the acceleration due to gravity d = 43 + 21t - 0.5at^2 Now substitute the known values into the formula. 5 = 43 + 21t - 0.5*9.8*t^2 Subtract 5 from both sides and perform the multiplication 0 = 38 + 21t - 4.9t^2 We now have a quadratic function with a=-4.9, b=21, c=38. Using the Quadratic formula, we can find the solutions at t = 5.66 and t=-1.37 Since the negative time doesn't make sense, the correct value will be t = 5.66 seconds.

1nba youngboy thxs for asking