THERE ARE 5 PROBLEMS

Question

THESE ARE THE 5 PROBLEMS 

Answer 1

Maximum height is reached when the first derivative is zero and second derivative is negative.

Ans 6) See here https://www.mathhomeworkanswers.org/279292/there-are-5-problems?show=279296#a279296

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h(t) = -16t^2 + 64t

h'(t) = -32t +64 = 0

or t = 2 seconds

h(2) = 64 feet

Ans 7) Maximum height = 64 feet

Ans 8) Time taken = 2 seconds

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f(t)= -5t^2 +10t +5

f'(t) = -10t +10 = 0

or t = 1 seconds

f(1) = 10 m

Ans 9) Maximum height above water is 10 m

Ans 10)  Time taken to reach maximum height  is 1 seconds

Answer 2

h(t)=-16t²+96t=

-16(t²-6t)=

-16(t²-6t+9-9)=

-16(t-3)²+144.

(1) Max height 144 feet.

(2) Time to reach 144 feet=3 seconds.

f(t)=-5t²+20t+5=

5-5(t²-4t+4-4)=

25-5(t-2)².

(3) Time to reach max height=2 seconds.

(4) Max height=25m.

h(t)=-16t²+160t+4=

4-16(t²-10t+25-25)=

404-16(t-5)².

(5) Time to reach maximum height=5 seconds.

(6) Maximum height = 404 feet.

h(t)=-16t²+64t=

-16(t²-4t+4-4)=

-16(t-2)²+64.

(7) Max height=64 feet.

(8) Time to reach maximum height = 2 seconds.

f(t)=-5t²+10t+5=

5-5(t²-2t+1-1)=

10-5(t-1)².

(9) Max height=10m.

(10) Time to reach max height=1 second.