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

a. 32 degrees

step-by-step explanation:

here we have to use the rule.

m∠e = 1/2 (major arc - minor arc)

m∠e = 1/2 (212 - 148)

m∠e = 1/2 (64)

m∠e = 32 degrees

answer: a. 32 degrees

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