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.

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