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L A ball is attached to a spring, which is stretched and then let go. The height of the ball is given by the sinusoidal equation y = -2.5cos(t) +6 %3D where y is the height above the ground in feet...

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L A ball is attached to a spring, which is stretched and then let go. The height of the ball is given by the sinusoidal equation y = -2.5cos(t) +6
%3D
where y is the height above the ground in feet andtis the number of seconds since the ball was released. How long, to the nearest tenth of a
second, does it take for the ball to return to its starting height?
Format
Font
Size
BIUSX, X })
x 99 A A
Ων
工に
事|三 = = =
Upload image n O
Extracted text: L A ball is attached to a spring, which is stretched and then let go. The height of the ball is given by the sinusoidal equation y = -2.5cos(t) +6 %3D where y is the height above the ground in feet andtis the number of seconds since the ball was released. How long, to the nearest tenth of a second, does it take for the ball to return to its starting height? Format Font Size BIUSX, X }) x 99 A A Ων 工に 事|三 = = = Upload image n O
Answered 128 days After Jun 06, 2022

Solution

Rajeswari answered on Oct 13 2022
60 Votes
112894 assignment
Initially i.e. at the start t=0
When t=0 height is given by
Y=-2.5 where t=0
Substitute t =0 . We know cos 0 = 1
Hence initial height
Y = -2.5+6 = 3.5 feet
The same height it would reach when again
3.5 =-2.5
Simplify to get
Cos 3pi/4 t = 1
This is possible when...
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