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?

Question

Hill

Physics

9 September, 22:49

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?

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Raine · 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.