Finding how long an object has been in the air for and the time took to reach max ht.

[theredeemer]

An object is launched at 13.8 m/s from the ground. The equation for the object's height "h" at time "t" seconds after launch is: h(t) = -4.9t² + 13.8t, where h is in meters.

 

[eumyang]

The object wouldn't be in the air anymore when it hits the ground, no? So what should you set h(t) equal to?

Assuming that you know nothing about calculus, recall that the vertex of a quadratic equation is the maximum or minimum point. So rewrite the equation in vertex form.

 

[VietDao29]

An object is launched at 13.8 m/s from the ground. The equation for the object's height "h" at time "t" seconds after launch is: h(t) = -4.9t² + 13.8t, where h is in meters.

There are a couple of ways of achieving this:

• The first way is to use your imagination. Notice that, when you launch the object from the ground, its speed is greatest, right? And then, as it flies off, the speed decreases. And when its speed is 0, it's also the time that object reaches its max height (before it starts to falls off).
  So, to find it's max height, we can calculate h(t₀), where t₀ is the time that its speed is 0.
  
• The second way is to try to Complete the Square to find the maximum value for h(t). Have you studied how to Complete the Square?
  In case you haven't, I'll explain it briefly so you can get the idea. Completing the Square is the manipulation you do to change your expression into something like this:
  
  (a + b)² + c, or c - (a + b)²​
  
  The identity we should use here is the Square of Sum: (a + b)² = a² + 2ab + b²
  
  Example: Complete the Square for:
  • $$x ^ 2 + 3x + 1$$
  • $$-x ^ 2 + 6x - 3$$
  
  ------------------------------
  Problem 1: $$x ^ 2 + 3x + 1$$.
  First we'll split it so that it contains the first 2 terms a² + 2ab, like this:
  $$x ^ 2 + 3x + 1 = x ^ 2 + 2.(x)\left(\frac{3}{2}\right) + 1$$
  Then, we'll add, and subtract b², note that, in this example: $$b = \frac{3}{2}$$ like this:
  $$... = x ^ 2 + 2.x\frac{3}{2} + 1 = x ^ 2 + 2.x\frac{3}{2} + \frac{9}{4} - \frac{9}{4} + 1 = \left(x ^ 2 + 2.x\frac{3}{2} + \frac{9}{4}\right) - \frac{5}{4}$$
  $$= \left( x + \frac{3}{2} \right) ^ 2 - \frac{5}{4}$$
  
  Problem 2: $$-x ^ 2 + 6x - 3$$.
  It's the same, except that we have a little minus sign "-", in front of x², right? So we can factor -1 out, like this, and do the same as what we've done above:
  $$-x ^ 2 + 6x - 3 = -(x ^ 2 - 6x + 3) = - (x ^ 2 + 2.x.3 + 3) = - (x ^ 2 + 2.x.3 + 9 - 9 + 3)$$
  $$= - [(x ^ 2 + 2.x.3 + 9) - 6] = -[(x + 3) ^ 2 - 6] = 6 - (x + 3) ^ 2$$
  
  After Completing the Square for h(t), you can find its maximum value by noting that: $$\alpha ^ 2 \ge 0 , \forall \alpha \in \mathbb{R}$$

Let's see if you can tackle this problem on your own. :)

 

[Redbelly98]

Moderator's note:

Since this thread is a duplicate, I am closing it.