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Visual Basic Compute Height After 3 Seconds
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Visual Basic Compute Height After 3 SecondsPosted:

-Kraddy
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This is the question:

Suppose a ball is thrown straight up in the air with an initial velocity of 50 feet per second and an initial height of 5 feet. How high will the ball be after 3 seconds? See Fig. 3.2. Note: The height after t seconds is given by the expression - 16 t 2 + v 0 t + h 0 , where v 0 is the initial velocity and h 0 is the initial height.


So far I have:


Public Class Projectilepg68

Private Sub btnOne_Click(sender As Object, e As EventArgs) Handles btnOne.Click
Dim initVel As Integer
Dim initHt As Integer
Dim finalHt As Integer
Dim sec As Integer

initVel = 50

initHt = 5

sec = 3

finalHt = -16 * sec ^ 2 + initVel * sec + initHt
End Sub

Private Sub txtOne_TextChanged(sender As Object, e As EventArgs) Handles txtOne.TextChanged

End Sub
End Class


When I run this, nothing comes out, I know I am missing something. Anyone care to help?
#2. Posted:
speed
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Nothing is coming out because you're not telling anything to come out. You're assigning the final value to finalHt, but doing nothing with it after that.

So to actually see the result, you'd want to do something like update a label's text, output to console, or something like that.
#3. Posted:
Skittle
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You also appear to have the TextChanged handler for txtOne but appear to be doing nothing with it.

As to what speed said, you need to output the result, as you are using a Windows form, you can simply use
MsgBox(finalHt)
to produce a message box, or one of the others that speed listed above.
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