Applying Solving Quadratics Notes
Launched/Falling Objects Problems
Remind students of the following formula and remind them about what the parts represent
H(t) = h0 + v0t – 16t2
Height = Initial Height + Initial Velocity*Time – 16*Time^2
The formula is quadratic, so a lot times, factoring can help us answer questions involving it
Example A cannon mounted on the top of 256ft tall fortress launches a shot in the air at an initial
velocity of 96ft/sec upward. How long will it take the cannonball to hit the ground?
H(t) = 256 + 96t – 16t2
We are looking for when it hits the ground, so H(t) = 0
0 = 256 + 96t – 16t2
Factor out -16 as the GCF and rearrange to standard form
-16(t2 – 6t – 16) = 0
Factor the trinomial Show the method of your choice
-16(t – 8)(t + 2) = 0
Solve for t
t = 8, -2
In the context of the problem, only 8 actually makes sense.
Our answer to the question is 8 seconds.
EX #1 A tennis player backhands the ball upper wards at a velocity of 28ft/sec. The player jumped
upward to hit the ball 8ft in the air. How long before the ball hits the ground?
0 = 8 + 28t – 16t2
The GCF here would be -4
-4(4t2 – 7t – 2) = 0
Review factoring or have the students try it
-4(4t + 1)(t – 2) = 0
Applying Solving Quadratics Notes
Solve each factor for t
t = -1/4, 2
Only 2 makes sense
The ball hits the ground after only 2 seconds
Ex# 2: A ground level super-slingshot launches a pumpkin upward at an initial velocity of 88ft/sec. At
what points in time will the pumpkin have a height of 120ft?
The big difference for this problem is that the initial height is zero, and the height we are looking for is
not 0
120 = 0 + 88t – 16t2
Rearrange and factor out a GCF of -8
-8(2t2 – 11t + 15) = 0
Factor the trinomial
-8(2t – 5)(t – 3) = 0
t = 5/2, 3
Both times are positive so we can make sense of both
At 2.5 and 3 seconds, the pumpkin will reach a height of 120ft
Applying area to quadratic problems
Example: A rectangle is enclosed by a fence with a total perimeter of 60yds. If the area of the rectangle
is 216yd2, then what are the dimensions of the rectangle?
Remind them of how we find both perimeter and area of a rectangle
L + W + L + W = Perimeter
2L + 2W = 60
LW = Area
LW = 216
Applying Solving Quadratics Notes
Solve for length in terms of width
2L = -2W + 60
L = -W + 30
Rewrite area using substitution
(-W + 30)(W) = 216
Guide the students through rewriting the quadratic and then solving. The -1 can factored out to make
the first term positive
-W2 + 30W – 216 = 0
-(W2 – 30W + 216) = 0
-(W – 12)(W – 18) = 0
Set each factor equal to 0 and solve
W = 12, 8
We can go back to find length, but the two dimensions end up being 12yd and 8yd
Ex # 3: A state is redrawing its counties into rectangular sections. Each section needs its borders to have
a total perimeter of 26miles. The area of each rectangle needs to be 42miles2. What should the
dimensions of the rectangle be?
Solve for L in terms of W using perimeter
L + W + L + W = 26
2L + 2W = 26
2L = -2W + 26
L = -W + 13
Use area and substitution to write a quadratic equation
LW = 42
(-W + 13)W = 42
-(W2 – 13W + 42)
Factor and solve the equation
Applying Solving Quadratics Notes
-(W – 6)(W – 7) = 0
W = 6, 7
So our dimensions for the counties are 6miles and 7miles