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WILL GIVE BRAINLIEST, THANKS AND 5 STARS! PLEASE HELP! LOTS OF POINTS
Sam kicks a ball off of a rooftop. After t seconds, its height, h (in feet), is given by the formula
h (t) = -16t2 + 64t + 80.
*NOTE: Complete this without using graphing features on your calculator, and show all work for credit. You may type out your work or upload it in the next question.
a. What is the maximum height reached by the ball?
b. When did the ball reach its maximum height?
c. When will the ball hit the ground? (Do algebraically)
d. What is the height of the ball after 1 second?
(THIS IS A 4 PART QUESTION, PLEASE DO ALL PARTS!!!) PLEASE GIVE INDEPTH EXPLANATION FOR ALL PARTS!

Sagot :

semsee45

Answer:

(a)  144 ft

(b)  2 s

(c)  5 s

(d)  128 ft

Step-by-step explanation:

[tex]h(t) = -16t^2 + 64t + 80[/tex]

Part (a)

To find the turning point of the parabola:

1. differentiate the function:

[tex]\implies h'(t) = -32t + 64[/tex]

2. Set it to zero and solve for t:

[tex]\implies h'(t) =0[/tex]

[tex]\implies -32t + 64=0[/tex]

[tex]\implies 32t = 64[/tex]

[tex]\implies t=2[/tex]

3. Input found value of t back into the function and solve for h:

[tex]\implies h(2) = -16(2)^2 + 64(2) + 80=144[/tex]

Therefore, the maximum height reached by the ball is 144 ft

Part (b)

As found in part (a), the maximum height is reached when t = 2, so the ball reached its maximum height at 2 seconds.

Part (c)

The ball will hit the ground when h(t) = 0

1. Set the function to zero

[tex]\implies h(t)=0[/tex]

[tex]\implies -16t^2 + 64t + 80=0[/tex]

2. Solve for t:

Divide both sides by 16

[tex]\implies -t^2 + 4t + 5=0[/tex]

[tex]\implies t^2 - 4t - 5=0[/tex]

Factor:

[tex]\implies t^2 +t-5t - 5=0[/tex]

[tex]\implies t(t+1)-5(t+1)=0[/tex]

[tex]\implies (t+1)(t-5)=0[/tex]

Therefore:

[tex](t+1)=0 \implies t=-1[/tex]

[tex](t-5)=0 \implies t=5[/tex]

As time is positive, t = 5

So the ball will hit the ground at 5 seconds.

Part (d)

To find the height of the ball after 1 second, input t = 1 into the function and solve for h:

[tex]\implies h(1) = -16(1)^2 + 64(1) + 80=128[/tex]

Therefore, the height of the ball after 1 second is 128 ft

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Parts (a) & (b) - alternative method

Rewrite the function in vertex form by completing the square.

Set the function to zero:

[tex]\implies-16t^2 + 64t + 80=0[/tex]

Subtract 144 from both sides:

[tex]\implies -16t^2+64t-64=-144[/tex]

Factor out -16:

[tex]\implies -16(t^2 - 4t +4)=-144[/tex]

Factor expression in brackets:

[tex]\implies -16(t-2)^2=-144[/tex]

Add 144 to both sides:

[tex]\implies -16(t-2)^2+144=0[/tex]

Therefore:

[tex]\implies h(t)=-16(t-2)^2+144[/tex]

The maximum height is the y-value of the vertex.

The vertex is (2, 144) so the maximum height is 144 ft.

The time the ball reached the max height is the x-value of vertex.

So the time then the ball reached the maximum height is 2 s

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