Height of Object Launched at 13.8 m/s

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  • #1
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.
 
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  • #2


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.
 
  • #3


theredeemer said:
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:
    • [tex]x ^ 2 + 3x + 1[/tex]
    • [tex]-x ^ 2 + 6x - 3[/tex]

    ------------------------------
    Problem 1: [tex]x ^ 2 + 3x + 1[/tex].
    First we'll split it so that it contains the first 2 terms a2 + 2ab, like this:
    [tex]x ^ 2 + 3x + 1 = x ^ 2 + 2.(x)\left(\frac{3}{2}\right) + 1[/tex]
    Then, we'll add, and subtract b2, note that, in this example: [tex]b = \frac{3}{2}[/tex] like this:
    [tex]... = 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}[/tex]
    [tex]= \left( x + \frac{3}{2} \right) ^ 2 - \frac{5}{4}[/tex]

    Problem 2: [tex]-x ^ 2 + 6x - 3[/tex].
    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:
    [tex]-x ^ 2 + 6x - 3 = -(x ^ 2 - 6x + 3) = - (x ^ 2 + 2.x.3 + 3) = - (x ^ 2 + 2.x.3 + 9 - 9 + 3)[/tex]
    [tex]= - [(x ^ 2 + 2.x.3 + 9) - 6] = -[(x + 3) ^ 2 - 6] = 6 - (x + 3) ^ 2[/tex]

    After Completing the Square for h(t), you can find its maximum value by noting that: [tex]\alpha ^ 2 \ge 0 , \forall \alpha \in \mathbb{R}[/tex]

Let's see if you can tackle this problem on your own. :)
 
  • #4


Moderator's note:

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

1. What is the formula for calculating the height of an object launched at 13.8 m/s?

The formula for calculating the height of an object launched at 13.8 m/s is h = (v2sin2θ)/2g, where h is the height, v is the initial velocity (13.8 m/s), θ is the angle of launch, and g is the acceleration due to gravity (9.8 m/s2).

2. How do you determine the angle of launch in the formula?

The angle of launch can be determined by using a protractor to measure the angle at which the object is launched from the ground. It is important to measure the angle accurately, as even a small difference can greatly affect the calculated height.

3. Is the initial velocity the only factor that affects the height of the object?

No, the height of the object is also affected by the angle of launch and the acceleration due to gravity. A higher initial velocity and a smaller angle of launch will result in a higher height, while a lower initial velocity and a larger angle of launch will result in a lower height. The acceleration due to gravity remains constant.

4. Can the formula be used for any object launched at 13.8 m/s?

Yes, the formula can be used for any object launched at 13.8 m/s, as long as the angle of launch and the acceleration due to gravity are known. However, it is important to note that the formula is derived for objects launched from a horizontal surface and may not be accurate for objects launched from other surfaces.

5. How can this formula be applied in real-life situations?

This formula can be applied in real-life situations such as calculating the height of a projectile launched from a catapult or a rocket launched into the air. It can also be used in sports such as basketball or baseball to determine the height of a ball thrown or hit at a certain velocity and angle. Additionally, this formula is used in engineering and physics to analyze the motion of objects.

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