Can some one please verify my work? Thank you!
Tom throws a baseball so that the height of the ball is given by h(t)= -5t^2 + 20t + 1 where h is the height reached in meters and t is the time taken in seconds.
a) If the ball is caught at the same height it is thrown, how long was it in the air?
What i did:
set h(t) = 1 because that is the height when we through the ball, and solve.
1= -5t^2 +20t + 1
0= -5t^2 + 20t
0= -5t(t - 4)
There fore t = 4 & 0, and the ball was in the air for 4 seconds.
b) IF the ball is caught at the same height at which it is thrown, what is the maximum height reached by the ball?
What I did:
We have two symmetrical pairs ( 0,1) and (4,1).
So the AOS:
x= (4+0)/2
x=2
Sub x=2 in the equation to solve for max. height!
h(2)= -5(2)^2+20(2)+1
=21
There fore the max height is 21.
In part (a) you set the initial height to 1, why did you not set the initial height to 0? Where 0 represents the position from the ball leaving Tom's hand. As I stated earlier, your mathematical work is sound.
You start off by subtracting 1 on both sides, to yield:
Then you factored out -5t, to yield:
Then you solved for t and got:
All that makes sense to me, what I don't get is the logic of setting the height to 1. For part (b), if you know the time of ball in the air then could you not calculate the height for half the total time?
What are you taking as the time the ball was initially thrown? Why are you assuming the ball was thrown from height 0? The word of the problem, "t is the time taken in seconds", implies that t measures the time it takes the ball to go from one person to the other- that is the ball was thrown when t= 0 so its height at the at time is -5(0^2)+ 20(0)+ 1= 1 meter as sakonpure6 used.
As I stated earlier, your mathematical work is sound.
You start off by subtracting 1 on both sides, to yield:
Then you factored out -5t, to yield:
Then you solved for t and got:
All that makes sense to me, what I don't get is the logic of setting the height to 1. For part (b), if you know the time of ball in the air then could you not calculate the height for half the total time?
Yes, this is correct. A parabola is symmetric about its vertex, which, in this case, is the greatest height of the ball. Another way to do this is to "complete the square". We can write [tex]h= -5t^2+ 20t+ 1= -5(t^2- 4t)+ 1= -5(t^2- 4t+ 4- 4)+ 1= -5(t^2- 4t+ 4)+ 20+ 1= -5(t^2- 4t+ 4)+ 21. The reason I added and subtracted 4, and then took the "-4" outside the parentheses, was to get a "perfect square" inside the parentheses: t^2- 4t+ 4= (t- 2)^2.
That gives h= -5(t- 2)^2+ 21. Since a square is never negative, -5(t- 2)^2 is never positive. At t= 2 (yes, halfway between 0 and 4), h= 21 meters. For any other t we have 21 minus something so that 21 meter is, in fact, the greatest height.
So the AOS:
x= (4+0)/2
x=2
Sub x=2 in the equation to solve for max. height!
h(2)= -5(2)^2+20(2)+1
=21
There fore the max height is 21.[/QUOTE]