Use the formula h = −16t2 + v0t. (if an answer does not exist, enter dne.) a ball is thrown straight upward at an initial speed of v0 = 56 ft/s. (a) when does the ball initially reach a height of 40 ft?
Using the given formula with v0=56 ft/s and h=40 ft h = -16t2 + v0t 40 = -16t2 + 56t 16t2 - 56t + 40 = 0 Solving the quadratic equation: t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32 We have two possible solutions t1 = (56+24)/32 = 2.5 t2 = (56-24)/32 = 1 So initially the ball reach a height of 40 ft in 1 second.
