Thread: Finding Max height and time

1. Finding Max height and time

The formula h = -16t^2 + v(subscript)0t + s(subscript)0 gives the height of an object tossed upward where v(subscript)0 represents the initial velocity, s(subscript)0 represents the initial height, and t represents time. A golf ball is hit straight up from the ground level with an initial velocity of 72 ft/sec. What is the maximum height that the ball reaches and the number of seconds it takes to reach that height?

2. Originally Posted by greenpumpkins
The formula h = -16t^2 + v(subscript)0t + s(subscript)0 gives the height of an object tossed upward where v(subscript)0 represents the initial velocity, s(subscript)0 represents the initial height, and t represents time. A golf ball is hit straight up from the ground level with an initial velocity of 72 ft/sec. What is the maximum height that the ball reaches and the number of seconds it takes to reach that height?
You have: $s_0 = 0$ and $v_0 = 72\ \frac{ft}{s}$
Plug in these values into the given equation:

$h(t) = -16t^2+72t$

This is the equation of a parabola opening down with it's maximum at it's vertex:

$h(t) = -16\left(t^2-\frac92 t+\frac{81}{16}\right)+81 = -16\left(t-\frac94\right)^2+81$

Therefore the coordinates of the vertex are $V\left(\frac94\ ,\ 81\right)$

That means after 2.25 s the golf ball reaches it's maximum height of 81 ft.