Baseball height question

Question

1. A baseball is hit from a height of 0.8 m. After t seconds, its height, h  meters, is given by the equation h=−5t^2+20t+0.8. How long does the ball stay at least 15 m above the ground?

Please answer in full detail

Answer 1

If h(t) = −5t² + 20t + 0.8 and you want to know when the height is at least 15 m then you want:

−5t² + 20t + 0.8 > 15 or −5t² + 20t - 14.2 > 0

To solve this inequality, pretend that it is an equality to find its roots (zeros):

−5t² + 20t - 14.2 = 0

Using the quadratic formula:

t = [-20 ± √(20² - 4(-5)(-14.2))]/(2(-5))

t = (-20 ± √116)/-10

t = (-20 ± 10.77)/-10

t = 0.92 or t = 3.08

For inequalities, these are the dividing points in which the value of the quadratic switches from positive to negative. Since this quadratic has a negative initial coefficient (-5), it goes up from the negative world at 0.92, stays positive until 3.08, then goes back negative. So the values of "t" where the quadratic is > zero are between 0.92 and 3.08 or a total of 3.08 - 0.92 = 2.2 seconds

So the ball is at least 15 meters above the ground for 2.2 seconds

Answer 2

h(t) = 15 = -5t^2 +20t + .8

5t^2 -20t +14.2 = 0

use the quadratic formula

t = 20/10 +/- (1/10)sqr(400-4(5)(14.2))

t =2 +/- 1.077 = about 3.077 and 0.923 seconds

from about 0.923 seconds to 3.077 seconds the ball is above 15 meters off the ground

max height is when t =20/10= 2 seconds

max height = -5(4)+20(2)+.8= 20.8 meters

at 13 meters, the time is less and more than 2 seconds by an equal amount

roughly it's above 15 meters from 1 second to 3 seconds

to 3 decimal places it's from 0.923 to 3.077 seconds above 15 meters