At what time does the ball reach its maximum height? Vertical Motion

Question

Ava and Rose are playing softball. Ava throws the ball underhand to Rose with an upward velocity of 35 ft per second and the ball leaves her hand 5 ft above the ground.

Use the formula: h(t)=-16t^2+Vot+ho

Answer 1

Hi Kennedy,

Without using concepts from Calculus, we can take advantage of the fact that we have a description of this ball's vertical position with respect to time:

h(t) = -16t²+v₀t + h₀

According to the problem, the initial height (h₀) of the ball is 5 ft, and its initial upward velocity (v₀) is 35 ft/s. Notice that this position function describes a downwards parabola. We remember from math class that the x coordinate of the vertex of a parabola y = ax² + bx + c is:

x = - b / 2a

In our case, the "x-axis" is actually the time axis and our parabola is h(t) = -16t²+35t + 5. So the vertex of this parabola (the maximum height the ball reaches) occurs when:

t = - b / 2a = - 35 / 2(-16) = 1.09 seconds