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Question
A ball is thrown in the air. Its height at any time t is given by h = 3 + 14t – 5t2. Find the maximum height it can reach.
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Solution
The height h at any t is given by h = 3 + 14t – 5t2
∴ `"dh"/dt = d/dt(3 + 14t - 5t^2)`
= 0 + 14 x 1 – 5 x 2t
= 14 – 10t
and `(d^2h)/(dt^2) = d/dt(14 - 10t)`
= 0 – 10 x 1
= – 10
The root of `"dh"/dt` = 0,
i.e. 14 – 10t = 0 is t = `(14)/(10) = (7)/(5)`
and
`((d^2h)/(dt^2))_("at" t = 7/5)` = −10 < 0
∴ By the second derivative test, h is maximum at t = `(7)/(5)`.
∴ Maximum height = `(3 + 14t – 5t^2)_("at" t = 7/5)`
= `3 + 14(7/5) - 5(7/5)^2`
= `3 + (98)/(5) - (245)/(25)`
= `(75 + 490 - 245)/(25)`
= `(320)/(25)`
= 12.8
Hence, the maximum height the ball can reach = 12.8 units.
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