Question

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

Answer 1

Given

Ladder height = 3 foot

Initisl Velocity = 45 ft/sec

`y=-16t^2+45t+6`

Find

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

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

Step-by-step explanation

We maximize y , to find critical point

`\begin{gathered} y^(\prime)=-32t+45 \\ t=(45)/(32) \end{gathered}`

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

Maximum Value is

`y((45)/(32))=-16*((45)/(32))+45*((45)/(32))+6=-31.640625+63.28125+6=37.640`

Final Answer

37.640