Use the model for trajectory, h=-16t^2+vt+h_0, with the velocity and initial height given.
B
b
Let h=10, and then solve for t. Make sure to set the quadratic equal to zero before factoring.
C
c
Consider that the velocity stays the same but was shot from a lower place. Think about how this will affect the time it takes to reach the ground again.
A
a
h=-16t^2+20t+6
B
b
1 sec
C
c
Raymond's ball will be in the air less time.
Explanation: See solution.
Practice makes perfect
a
To write an equation for Jerald's throw, we need to use the vertical projectile model.
h= -16t^2+ vt+ h_0
We can replace v with the velocity of 20 ft/sec and h_0 with the initial height of 6 ft.
h= -16t^2+ 20t+ 6
b
To find how long it takes for the ball to reach the hoop, we need to use our equation from Part A and set h equal to the height of the hoop, 10 ft.
10 = -16t^2+20t+6
Before we can factor the quadratic, we need to set it equal to zero. Let's move everything to the left hand side of the equation.
Now that the equation is equal to zero, we can factor. First let's factor out the greatest common factor, 4.
4(4t^2-5t+1) = 0
We now know that 4t^2-5t+1=0. In this quadratic, a=4, b=-5, and c=1.
4t^2-5t+1=0
⇕
ax^2+bx+c=0
⇕
a=4, b=-5, c=1
Let's take the product ac and then look at which factors add up to b. In this case, ac = 4. Since the product is positive, and b is negative, both factors will be negative.
Factors of 4
Sum of Factors
-1, -4
-5
-2, -2
-4
Since -1, -4 have the sum of -5, we can replace the linear term, -5t= -4t+- t.
4t^2 -5t+1 = 4t^2 -4t -t +1
Let's factor.
The ball will be at 10 ft once on it's way up after 0.15 sec and a second time on its way down at 1 sec. In the case of when it reaches the hoop, if the shot is good, then it will be on the way down after 1 sec.
c
Here, we need to consider what is different about Raymond's shot from Jerald's. Let's have a look at how these trajectories might look when launched from different heights.
It looks like they land at the same time. However, if we take a closer look at the place where the two curves cross the x -axis, we can see one lands slightly before the other.
We can see from the first graph that Jerald's throw, the one that starts at 6 ft, will have a higher maximum height compared to Raymond's. It makes sense that the ball thrown from the shorter height has a shorter amount of time in the air given they have the same velocity.
