Question

A ball is thrown from an initial height of 3 meters with an initial upward velocity of 30(m)/(s). The ball's height h (in meters) after t seconds is given by the following.

h=3+30t-5t^(2)
Find all values of t for which the ball's height is 13 meters.

Answer 1

Answer: -451

Explanation:

so, the height we are trying to find is 13 meters.

13=3+30t-5t^2

so, we can go ahead and pull that number in.

now we have to subtract the 3 to try and get as much as possible over on the other side to find t.

10=30t-5t^2

Answer 2

Explanation:

h(t) = 3 + 30t - 5t²

the height is set to 13 meters.

since the ball will go up and then come back down again, we expect 2 points of time, when the ball is at 13 meters high. if any at all, by the way (keep that in mind for similar questions, the object might not even reach a certain height).

13 = 3 + 30t - 5t²

10 = 30t - 5t²

2 = 6t - t²

0 = -t² + 6t - 2

the general solution to a quadratic equation

ax² + bx + c = 0

is

x = (-b ±sqrt(b² - 4ac))/(2a)

in our case

x = t

a = -1

b = 6

c = -2

t = (-6 ±sqrt(6² - 4×-1×-2))/(2×-1) =

= (-6 ±sqrt(36 - 8))/-2 = (-6 ±sqrt(28))/-2 =

= (-6 ±sqrt(4×7))/-2 = (-6 ±2×sqrt(7))/-2 =

= 3 ± sqrt(7)

t1 = 3 + sqrt(7) = 5.645751311... seconds

t2 = 3 - sqrt(7) = 0.354248689... seconds

on its way up the ball will pass the mark of 13 meters after about 0.35 seconds, and on its way back down it will pass the mark of 13 meters again after about 5.65 seconds.