A grapefruit is tossed straight up with an initial velocity of $50 \mathrm{ft} / \mathrm{sec}$. The grapefruit is 4 feet above the ground when it is released. Its height at time $t$ is given by
\[
y=-16 t^{2}+50 t+4
\]
How high does it go before returning to the ground?
Round your answer to one decimal place.
The maximum height of the grapefruit is feet.

Answer

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Answer

Round the maximum height to one decimal place: $\boxed{43.1}$ feet

Steps

Step 1 ：Given the equation for the height of the grapefruit: $y = -16t^2 + 50t + 4$

Step 2 ：Find the vertex of the parabola: $t_{vertex} = \frac{-b}{2a}$

Step 3 ：Substitute the values of a and b: $t_{vertex} = \frac{-50}{2(-16)} = 1.5625$

Step 4 ：Find the maximum height by substituting the value of $t_{vertex}$ into the equation: $y_{vertex} = -16(1.5625)^2 + 50(1.5625) + 4 = 43.0625$

Step 5 ：Round the maximum height to one decimal place: $\boxed{43.1}$ feet