Frame 2 or 3 real life-based equations and solve them.
Answers
Answer:
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.