The arrow was launched vertically upward with an initial speed of 30 m / s. At what point in time will the boom reach its maximum height?

When the body moves up, we take the height formulas h = (V ^ 2-Vo ^ 2) / – 2g, where V is the final speed, Vo is the initial = 0, since at the highest point the speed is zero, g is the acceleration of gravity = 10m / s ^ 2. Hence h = 45m. Then we take the formula h = Vot-gt ^ 2/2, substitute the known values and height, we get the quadratic equation 5t ^ 2-30t + 45, whence t = 3c.