Finding Max height and time

**greenpumpkins** · October 26th 2008, 06:52 PM

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?

**earboth** · October 27th 2008, 12:53 AM

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.

Thread: Finding Max height and time

LinkBack

Thread Tools

Display

Finding Max height and time

Similar Math Help Forum Discussions

Height above ground at time t, of ferris wheel

height, time, speed

[SOLVED] Polynomial time/height

Function of height and time?

Hi first time on height question

Search Tags