Algebra 2 Gravity Word Problems
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What Goes Up
Gravity Word Problems


 When you throw something into the air,

 it goes up (probably).
 As soon as it leaves your hand, gravity starts slowing it down.
 At some point it stops going up and starts coming back down.
 This equation tells us how high IN FEET the thing is in the air
 at some time after it is thrown.

 Height = -16t2 + Vot + ho

 Now before you can do anything with this mess 
 you need to know what all of those letters mean.
 t = time in seconds since you let the thing go
 Vo = the velocity (speed) the thing was going when it left your hand
 ho = how high your hand was off the ground when you let go of the thing
 Height = how high the thing is at any time "t."
 In this problem, we have t instead of X as the independent variable.
 The Algebra still all works he same way.
 If the t really bothers you, you can use X instead.
 A ball is thrown upward at a speed of 50 feet per second 
 (that's the Vo part)
 It is 6 feet in the air when it leaves the person's hand
 (that's the ho part)
 How high above the person's hand does it get???
 OK, let's see what numbers we know:
 t = it's the unknown, it's an independent variable (like X).
 Vo = 50 feet per second
 ho = 6 feet
 Height = another unknown. It depends on t, so it's a dependent variable (like Y)
 So when we put the numbers we know into the equation we get:

 Height = -16t2 + 50t + 6

 We work this problem exactly the same way as if it was Y = -16X 2 + 50X + 6
 Since the number next to the squared variable is negative (-16),
 the parabola opens down.


 So here we have:

 Height = -16t2 + 50t + 6


 A = -16    B = 50   C = 6

 The vertex (top) of the parabola happens at:


 About a second and a half ...

 Now we can use that number to see how high it is at that time.


 Now here's the tricky part.
 The question DID NOT ask how high the ball was in the air.
 It asked how high the ball was ABOVE THE PERSON'S HAND.
 The person's hand is 6 feet in the air,
 so the ball is about 39 feet above the person's hand!
 There is another thing that you might get asked
 on a problem like this.
 That is, how long until the ball hits the ground?
 To solve that, 
 we need to find the time when the height of the ball equals zero.
 That is, when it lands.

 Height = -16t2 + 50t + 6


  0 = -16t2 + 50t + 6

 Now we need the quadratic formula ...


 And here

 A = -16    B = 50   C = 6



 Saying that a ball hit the ground at -.11 seconds 
 doesn't have much meaning here so we can ignore that one.
 Saying that the ball hit the ground after 3.24 seconds does mean something,
 so that's our answer.
 So just where did that -.11 seconds come from anyway?
 Here's where ...
 The shape of the graph is an upside down parabola.
 That is also what the flight path of the ball looks like.


 But a parabola really doesn't start or end anywhere,
 it goes on forever.
 That means the parabola will cross the "ground" in two places.


 And that's where the other answer comes from.

   copyright 2005 Bruce Kirkpatrick

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