Question

FIFTY POINTS

y=-16t^2+40t+2

For what interval or intervals of time will the projectile be
above 18 feet?
a.) Between 1 and 1.5 seconds
b.) Less than 1 second and greater than 1.5 seconds
c.) Between 0.5 and 2 seconds
d.) Less than 0.5 second and greater than 2 seconds

Answer 1

To find when the projectile is above 18 feet, we need to find the values of time when the height of the projectile is greater than 18. The height of the projectile can be found by plugging the value of time into the equation y = -16t^2 + 40t + 2.

So, we need to find the values of t when y > 18.

y = -16t^2 + 40t + 2 > 18

-16t^2 + 40t - 16 > 0

To solve this inequality, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

where a = -16, b = 40, and c = -16.

So, t = (40 ± √(40^2 - 4(-16)(-16))) / 2(-16)

t = (40 ± √(40^2 + 256)) / -32

t = (40 ± √(1616)) / -32

t = (40 ± 40) / -32

t = (80 / -32) / 2 or (0) / 2

t = -1.25 or 0

Since t has to be positive (time cannot be negative), we can only use the positive value, t = 0.5. So, the interval of time when the projectile is above 18 feet is between 0.5 and 2 seconds, which is (c).

Answer 2

Check out the attached photo