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Sagot :
Given:
The speed is 64 feet per second.
The height of the high cliff is 80 feet.
The function is
[tex]H(t)=-16t^2+64t+80[/tex]a)
We need to find the maximum value of t in the given function to find a time when the rock reaches its maximum height.
Differentiate the given equation, we get
[tex]H^{\prime}(t)=-16(2t)^{}+64[/tex][tex]H^{\prime}(t)=-32t^{}+64[/tex]Set H'(t)=0 and solve for t.
[tex]0=-32t^{}+64[/tex]Adding 32t on both sides, we get
[tex]0+32t^{}=-32t+64+32t[/tex][tex]32t^{}=64[/tex]Dividing both sides by 32, we get
[tex]\frac{32t}{32}^{}=\frac{64}{32}[/tex][tex]t=2[/tex]Hence the rock reaches its maximum height after 2 seconds.
b)
Substitute t=2 in the given equation to find the maximum height of the rock.
[tex]H(2)=-16(2)^2+64(2)+80[/tex][tex]H(2)=144[/tex]Hence the maximum height obtained by the rock is 144 feet above sea level.
c)
Substitute H(t)=0 in the given function to find the time when the rock hit the ocean.
[tex]0=-16t^2+64t+80[/tex]Dividing both sides by (-16), we get
[tex]0=-\frac{16t^2}{-16}+\frac{64t}{-16}+\frac{80}{-16}[/tex][tex]0=t^2-4t-5[/tex][tex]t^2-4t-5=0[/tex][tex]t^2+t-5t-5=0[/tex][tex]t(t+1)-5(t+1)=0[/tex][tex](t+1)(t-5)=0[/tex][tex](t+1)=0,(t-5)=0[/tex][tex]t=-1,t=5[/tex]Omitting the negative value, we get t= 5 seconds.
Hence the rock hits the ocean after 5 seconds.
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