Question

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?

Answer 1

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.