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Sagot :
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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