show the following in the form of pictures of balls
1)=-3*4
2)=2*-4
3)=-3*-4
4)=4*5
play with balls
Answers
Answer:
Example: Throwing a Ball
A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. When does it hit the ground?
Ignoring air resistance, we can work out its height by adding up these three things:
(Note: t is time in seconds)
The height starts at 3 m: 3
It travels upwards at 14 meters per second (14 m/s): 14t
Gravity pulls it down, changing its position by about 5 m per second squared: −5t2
(Note for the enthusiastic: the -5t2 is simplified from -(½)at2 with a=9.8 m/s2)
Add them up and the height h at any time t is:
h = 3 + 14t − 5t2
And the ball will hit the ground when the height is zero:
3 + 14t − 5t2 = 0
Which is a Quadratic Equation !
In "Standard Form" it looks like:
−5t2 + 14t + 3 = 0
It looks even better when we multiply all terms by −1:
5t2 − 14t − 3 = 0
Let us solve it ...
There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give a×c, and add to give b" method in Factoring Quadratics:
a×c = −15, and b = −14.
The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15
By trying a few combinations we find that −15 and 1 work (−15×1 = −15, and −15+1 = −14)
Rewrite middle with −15 and 1: 5t2 − 15t + t − 3 = 0
Factor first two and last two: 5t(t − 3) + 1(t − 3) = 0
Common Factor is (t − 3): (5t + 1)(t − 3) = 0
And the two solutions are: 5t + 1 = 0 or t − 3 = 0
t = −0.2 or t = 3
The "t = −0.2" is a negative time, impossible in our case.
The "t = 3" is the answer we want:
The ball hits the ground after 3 seconds!
quadratic graph ball
Here is the graph of the Parabola h = −5t2 + 14t + 3
It shows you the height of the ball vs time
Some interesting points:
(0,3) When t=0 (at the start) the ball is at 3 m
(−0.2,0) says that −0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.
(3,0) says that at 3 seconds the ball is at ground level.
Also notice that the ball goes nearly 13 meters high.
Note: You can find exactly where the top point is!
The method is explained in Graphing Quadratic Equations, and has two steps:
Find where (along the horizontal axis) the top occurs using −b/2a:
t = −b/2a = −(−14)/(2 × 5) = 14/10 = 1.4 seconds
Then find the height using that value (1.4)
h = −5t2 + 14t + 3 = −5(1.4)2 + 14 × 1.4 + 3 = 12.8 meters
So the ball reaches the highest point of 12.8 meters after 1.4 seconds.
Answer:
1} -12
2}-8
3}-12
4}20
I hope this right