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While standing on a 3-foot ladder, a grapefruit is tossed straight up with an initial velocity of 45 ft/sec. Theinitial position of the grapefruit is 6 feet above the ground when it is released. Its height at time t is given byy = h(t) = -16t^2 + 45t + 6.a) How high does it go before returning to the ground? Round time to 2 decimal places to compute height.____feet.b) How long does it take the grapefruit to hit the ground? Round time to 3 decimal places.seconds

Respuesta :

Given

Ladder height = 3 foot

Initisl Velocity = 45 ft/sec

[tex]y=-16t^2+45t+6[/tex]

Find

(a) How high it go before returning to the ground.

(b) How long does it take the grapefruit to hit the ground.

Explanation

We maximize y , to find critical point

[tex]\begin{gathered} y^{\prime}=-32t+45 \\ t=\frac{45}{32} \end{gathered}[/tex]

It is a maxima as the given curve is a downward facing parabola.

Maximum Value is

[tex]y(\frac{45}{32})=-16\times(\frac{45}{32})+45\times(\frac{45}{32})+6=-31.640625+63.28125+6=37.640[/tex]

Final Answer

37.640

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