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It is known that, near the surface of the earth, a body falls with acceleration 2 (due to gravity) of about 32 ft/sec If we let h(t) be the height of the object at time t (measured in seconds), then our information is that h (t) = 32 Observe the minus sign to indicate that height is decreasing Now we will do some organized guessing What could h be It is some function whose derivative is the constant 32 Our experience indicates that polynomials decrease in degree when we differentiate them That is, the degree goes from 5 to 4, or from 3 to 2 Since, h is a polynomial of degree 0, we therefore determine that h will be a polynomial of degree 1 A moment s thought then suggests that h (t) = 32t This works! If h (t) = 32t then h (t) = [h (t)] = 32 In fact we can do a bit better Since constants differentiate to zero, our best guess of what the velocity should be is h (t) = 32t + v0 , where v0 is an undetermined constant Now let us guess what form h(t) should have We can learn from our experience in the last paragraph The antiderivative of 32t (a polynomial of degree 1) should be a polynomial of degree 2 After a little fiddling, we guess 16t 2 And this works The antiderivative of v0 (a polynomial of degree 0) should be a polynomial of degree 1 After a little fiddling, we guess v0 t And this works Taking all this information together, we find that the antiderivative of h (t) = 32t + v0 is h(t) = 16t 2 + v0 t + h0 ( )

Description Alert (bell) Backspace Form feed New line (linefeed) Carriage return Horizontal tab Vertical tab Null Single quote Double quote Backslash

Part I:

C# supports one other type of literal: the string A string literal is a set of characters enclosed by double quotes For example,

Notice that we have once again thrown in an additive constant h0 This does no harm: h (t) = [ 16t 2 ] + [v0 t] + [h0 ] = 32t + v0 , just as we wish And, to repeat what we have already confirmed, h (t) = [h (t)] = [ 32t] + [v0 ] = 32 We now have a general formula (namely ( )) for the position of a falling body at time t (Recall that we were first introduced to this formula in Section 26) See Figure 317 Before doing some examples, we observe that a falling body will have initial velocity 0 Thus 0 = h (0) = 32 0 + v0 Hence, for a falling body, v0 = 0 In some problems we may give the body an initial push, and then v0 will not be zero

"this is a test"

is a string You have seen examples of strings in many of the WriteLine( ) statements in the preceding sample programs In addition to normal characters, a string literal can also contain one or more of the escape sequences just described For example, consider the following program It uses the \n and \t escape sequences

Hi Jim: I am writing in response to yesterday s conference call in which I was embarrassed that I wasn t up-to-date in my information about the ABC Corporation deal I think it s important that we address this issue right away First, ABC said that I need to coordinate with you more effectively to save them the job of repeating conversations This was pretty humiliating to me as you and I certainly ought to be in close touch about this customer Second, I expect you to keep me up-to-date at every step along the way I expect you to do that without my having to hear updates from the customer Best, Paula

// Demonstrate escape sequences in strings using System; class StrDemo { static void Main() { ConsoleWriteLine("Line One\nLine Two\nLine Three"); ConsoleWriteLine("One\tTwo\tThree"); ConsoleWriteLine("Four\tFive\tSix"); // Embed quotes ConsoleWriteLine("\"Why \", he asked"); } }

The output is shown here:

Suppose that a falling body hits the ground with velocity 100 ft/sec What was the initial height of the body