Projectile Motion

Question

A projectile is thrown upward so that its distance above the ground after t seconds is h = -15t^2 + 510t. What is its maximum height?

Answer 1

h = -15t² + 510t is a quadratic equation and the graph of the quadratic, a parabola, is inverted resulting in a maximum value. To find the maximum value, find the vertex. To find the vertex, use the "cheat" t = -b/(2a) where a = -15 and b = 510.

t_vertex = -b/(2a)

t_vertex = -510/(2(-15))

t_vertex = -510/(-30)

t_vertex = 17

Now, to find the height, plug in t = 17

h = -15t² + 510t

h = -15(17)² + 510(17)

h = -15(289) + 8670

h = -4335 + 8670

h = 4335 units high

Answer 2

h' = - 30t + 510 = 0; t = 17 s

Max h = h(17) = - 15·17² + 510·17 = 4335 (unith of length)