Answer:
Step-by-step explanation:
[tex]y=-2sin (\frac{2\pi (t+1)} {7})+5\\Height ~is~maximum~\\if sin (\frac{2\pi (t+1)}{7} )=-1\\or~ \frac{2\pi (t+1)}{7} =\frac{3 \pi }{2} \\t+1=\frac{3\pi }{2} \times \frac{7}{2\pi } =\frac{21}{4} \\t=\frac{21}{4} -1=\frac{17}{4} =4.25 ~seconds.[/tex]
The weight reaches its maximum height at 4.25 seconds.
'The maximum height of the object is the highest vertical position along its trajectory.'
According to the given problem,
y = -2sin ([tex]\frac{2\pi ( t + 1 )}{7}[/tex]) + 5
For maximum height,
sin ([tex]\frac{2\pi ( t + 1 )}{7}[/tex]) = -1
⇒ [tex]\frac{2\pi ( t + 1 )}{7}[/tex] = [tex]sin^{-1} (-1)[/tex]
⇒ [tex]\frac{2\pi ( t + 1 )}{7}[/tex] = [tex]\frac{3\pi }{2}[/tex]
⇒ [tex]t +1=\frac{7 * 3\pi }{2 *2\pi }[/tex]
⇒ [tex]t +1[/tex] = [tex]\frac{21}{4}[/tex]
⇒ [tex]t = \frac{21}{4} -1[/tex]
⇒ t = [tex]\frac{17}{4}[/tex]
⇒ t = [tex]4.25[/tex] seconds.
Hence, we can conclude that at t = 4.25 seconds, the weight reaches its maximum height.
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