When a ball is thrown, its height in feet h after t seconds is given by the equation?

Question

h = vt - 16t^2

where v is the initial upwards velocity in feet per second. If v = 19 feet per second, find all values of t for which h = 5 feet. Do not round any intermediate steps. Round your answer to two decimal places.

Answer 1

h = vt - 16t^2
Since v=19 feet/sec and h=5 ft,
5 = 19t - 16t^2

16t^2 - 19t + 5 = 0

Solving the quadratic equation,

t1 = (-19 + sqrt(19^2 -4*16*5))/(2*19) and
t2 = (-19 - sqrt(19^2 -4*16*5))/(2*19)

In other words, t1 = 0.79 second and t2 = 0.39 second are the two values of t for which h=5 feet.

Answer 2

h = vt - 16t^2
Substituting v = 19 and h = 5 gives:
5 = 19t - 16t^2
Therefore, 16t^2 -19t + 5 = 0
Using the quadratic formula gives:
t = {19 ± sqrt[19^2 - 4(16)(5)]} / (2*16)
= [19 ± sqrt(41)] / 32
= 0.39 or 0.79 seconds