Aball is shot from a Cannon with up ward velocity of 40ft/sec the height is given by h(t)=-16tsquared+40t+ 5 find the maximum height

The height of the ball is given by the quadratic function:

$$h(t) = -16t^2 + 40t + 1.5$$

This is a parabola opening downwards (since the coefficient of $t^2$ is negative), so its vertex represents the maximum height.

For a quadratic function in the form $at^2 + bt + c$, the time at which the maximum (or minimum) occurs is given by the formula $t = -\frac{b}{2a}$.
In this equation, $a = -16$ and $b = 40$.

$$t = -\frac{40}{2(-16)}$$ $$t = -\frac{40}{-32}$$ $$t = \frac{40}{32}$$ $$t = \frac{5}{4} \text{ seconds}$$ $$t = 1.25 \text{ seconds}$$

Substitute $t = \frac{5}{4}$ into $h(t)$:

$$h\left(\frac{5}{4}\right) = -16\left(\frac{5}{4}\right)^2 + 40\left(\frac{5}{4}\right) + 1.5$$ $$h\left(\frac{5}{4}\right) = -16\left(\frac{25}{16}\right) + \frac{200}{4} + 1.5$$ $$h\left(\frac{5}{4}\right) = -25 + 50 + 1.5$$ $$h\left(\frac{5}{4}\right) = 25 + 1.5$$ $$h\left(\frac{5}{4}\right) = 26.5$$

The maximum height is 26.5 feet.

The maximum height is $26.5 \text{ ft}$.