Find Max Height & Time of Object Launched at 13.8m/s

[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. a. Using the MIN/MAX feature to find the time required for the object to reach it's maximum height. (How do you find it with the features? I know how to use it, but what do you do when you find it?)
b. Maximum height of object? (I got 9.7M)
c. How long is object in air?

RE: I guess you don't really need a graphics calculator. I'm only in Advanced Algebra currently at the moment, and some problems given to us involves some physics and chemistry problems.

 

[gneill]

A little algebra and calculus solves the questions easily. No graphics calculator required.

So, is your question about how to use your calculator, or how to solve he problem?

 

[theredeemer]

What I would like to know is how to solve and get the answer to 1 and 3. I got #2.

 

[jhae2.718]

Based on the "REQUIRES GRAPHICS CALCULATOR", I'm assuming you haven't yet taken a calculus course? (The answer to this question effects the explanation.)

 

[Squall]

Well if you plan on using your graphing calculator: you simply graph the function and use the calculators min/max function to find the x value at which it reaches maximum height, which corresponds to the peak of the parabola. Or if you know some calculus you can always take the derivative of the function and set it equal to zero and solve for t. Either way you should get the answer for part (a.
And part c should not be any harder using your calculator or projectile motion equations.

 

[jhae2.718]

For part c, what is the height of the projectile once it is no longer in the air? What time would this occur at? Part C requires only the solution to a quadratic.

If you're using a TI, the maximum function (2nd > Calc > maximum) should display the x and y values of the function at the maximum, which correspond to t and h.

In case you're interested, here's a brief explanation of the calculus involved:

Spoiler

To find the maximum or minimum value of a curve. You need to take the derivative. The derivative measures change over time. You can think of it as the slope extended to more complicated graphs than lines. (The derivative of a line is its slope.)

For a power function, given by xⁿ, the derivative is n*x^n-1. So, for your function h(t)=-4.9t²+13.8t, we look at each term separately.

The derivative of h(t), called h'(t) or h prime, is: -4.9*2t^2-1+13.8t^1-1, or h(t)=-9.8t+13.8

Set this equal to zero and solve for t and you have the answer for part A. Plug your answer for A into h(t), and you get the answer to part B.

 

[theredeemer]

I have no idea how to do problem a with the graphics calculator. When you use max, where are you supposed to move the target to find the time?

 

[jhae2.718]

For a TI-83/84: enter the equation into y=, graph it, then go to 2nd->Calc->Maximum. Put your lower bound to the left of the tallest part of the graph and your upper bound to the right of the tallest point. Then put your guess near the top.

The calculator will then display a value for x and a value for y on the screen. x in this case is t, the answer to A, and y is h, the answer to B.

 

[gneill]

I've never owned a 'graphics calculator', so I can't help you to use it to find the answer. But I can tell you that it took me about 10 seconds to use a bit of calculus to find the time required for the object to reach it's maximum height!

 

[theredeemer]

What are you supposed to plug in now to get c?

 

[jhae2.718]

Set h=0 and solve for t.

 

[ashishsinghal]

you have got the height now just put it in place of h(t) in your equation and solve