The height, h(t), of a baseball, in metres, at time t seconds after it is tossed out of a window is modelled by the function h(t) = -5t^2 + 20t + 15. A boy shoots at the baseball with a paintball gun. The
trajectory of the paintball is given by the function g(t) = 3t + 3. Will the paintball hit the baseball? If so, when? At what height will the baseball be?
Well, it seems like we have a real-life action movie going on here! Let's find out if the paintball hits the baseball.
To determine if the paintball hits the baseball, we need to find the point where the height of the baseball and the height of the paintball are the same.
The height of the baseball is given by h(t) = -5t^2 + 20t + 15, and the height of the paintball is given by g(t) = 3t + 3.
So, we need to solve the equation -5t^2 + 20t + 15 = 3t + 3.
Simplifying, we get -5t^2 + 17t + 12 = 0.
Now, we can factor or use the quadratic formula to find the values of t when the paintball hits the baseball. But hey, let's go with some humor and use the quadratic formula because it's much more entertaining:
The quadratic formula states:
t = (-b ± √(b^2 - 4ac))/(2a)
In our case, a = -5, b = 17, and c = 12. Plugging these values in, we get:
t = (-17 ± √((17)^2 - 4*(-5)*(12)))/(2*(-5))
Now, let's calculate the value under the square root:
(17)^2 - 4*(-5)*(12) = 289 + 240 = 529
Taking the square root of 529, we get 23. So:
t = (-17 ± 23)/(2*(-5))
Calculating both possibilities individually:
When t = (-17 + 23)/(2*(-5)) = 6/(-10) = -0.6
When t = (-17 - 23)/(2*(-5)) = -40/(-10) = 4
Now, let's analyze the results:
Since time can't be negative, we can ignore the first solution t = -0.6.
The paintball and baseball collide at t = 4 seconds.
To find the height of the baseball at that time, we can simply substitute t = 4 into the equation for h(t):
h(4) = -5(4)^2 + 20(4) + 15
= -80 + 80 + 15
= 15 meters
So, the baseball will be at a height of 15 meters when the paintball hits it. Keep in mind that this is just a mathematical model, so in reality, please don't shoot paintballs at baseballs out of windows!
To determine if the paintball will hit the baseball, we need to find the time when the height of the baseball and the height of the paintball are equal.
Since the height of the baseball is given by h(t) = -5t^2 + 20t + 15 and the height of the paintball is given by g(t) = 3t + 3, we can set these two equations equal to each other:
-5t^2 + 20t + 15 = 3t + 3
Now we can solve for t by rearranging the equation and simplifying:
-5t^2 + 20t + 15 - 3t - 3 = 0
-5t^2 + 17t + 12 = 0
This is a quadratic equation. We can solve it by factoring or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -5, b = 17, and c = 12. Plugging in these values, we have:
t = (-17 ± √(17^2 - 4(-5)(12))) / (2(-5))
Simplifying further:
t = (-17 ± √(289 + 240)) / (-10)
t = (-17 ± √(529)) / (-10)
t = (-17 ± 23) / (-10)
This gives two possible solutions for t:
t1 = (-17 + 23) / (-10) = 0.6
t2 = (-17 - 23) / (-10) = 4
Now we need to determine if the paintball hits the baseball at these times.
Let's substitute these values of t back into the equations:
At time t1 = 0.6 seconds:
h(t1) = -5(0.6)^2 + 20(0.6) + 15 = 13.8
At time t2 = 4 seconds:
h(t2) = -5(4)^2 + 20(4) + 15 = 35
So, the paintball hits the baseball at t = 0.6 seconds, and the height of the baseball at that time is 13.8 meters.
To determine if the paintball hits the baseball, we need to find the time(s) when the heights of the baseball and paintball are equal.
Let's set the heights of the baseball and paintball equal to each other:
h(t) = g(t)
Substituting the given functions:
-5t^2 + 20t + 15 = 3t + 3
To solve this equation, we need to rearrange it to the standard quadratic form:
-5t^2 + 20t - 3t + 15 - 3 = 0
-5t^2 + 17t + 12 = 0
We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
Given a = -5, b = 17, and c = 12:
t = (-17 ± sqrt(17^2 - 4(-5)(12))) / (2*(-5))
Simplifying this expression:
t = (-17 ± sqrt(289 + 240)) / (-10)
t = (-17 ± sqrt(529)) / (-10)
t = (-17 ± 23) / (-10)
So we have two possible values for t:
t1 = (-17 + 23) / (-10) = 6 / (-10) = -0.6
t2 = (-17 - 23) / (-10) = -40 / (-10) = 4
Since time cannot be negative, we discard the value t1 = -0.6.
Therefore, the paintball will hit the baseball at t = 4 seconds.
To find the height of the baseball at that time, substitute t = 4 into the height function:
h(4) = -5(4)^2 + 20(4) + 15
h(4) = -5(16) + 80 + 15
h(4) = -80 + 80 + 15
h(4) = 15
So at t = 4 seconds, the height of the baseball is 15 meters.
You'll be solving
-5t^2 + 20t + 15 = 3t + 3
5t^2 - 17t - 12 = 0
(t - 4)(5t + 3) = 0
t = 4 or t = -3/5, but let's assume t ≥ 0
They will hit after 4 seconds,
to find the height, sub t = 4 into h(t)