An object is launched from ground directly upward. The height (in metres) of the object after time t (in seconds) is given by h=-4.9t^2 + 39.2t (a) When will the object fall back to the ground? (b) Find the maximum height the object reaches. (c) For what values of t the height formula is applicable?

see below

Step-by-step explanation:

h=-4.9t^2 + 39.2t

When will if fall back to the ground

When will h=0

0 = -4.9t^2 + 39.2t

Factor out a - t

0 = - t (4.9t - 39.2)

Using the zero product property

-t = 0 4.9t - 39.2 = 0

t = 0 4.9t = 39.2

t = 39.2 / 4.9

t = 8

It will reach the ground again at 8 seconds

The maximum height is halfway between the zeros

(0+8) / 2 = 4

It will reach the max height at 4 seconds, substituting this in to find the max height

h=-4.9 (4) ^2 + 39.2 (4)

h = 78.4 meters

This will only be applicable between t=0 and t=8 seconds

0≤t≤8 seconds