The height h(t), of a baseball, in meters, at time t seconds after it is tossed out of a window is
modelled by the function h(t) = −5t
2 + 20t + 15.
a) What is the maximum height that the baseball can reach?
b) At what height is the window?
c) At what time will the ball hit the ground?
d) What is the axis of symmetry of this function?
h(t) = −5t^2 + 20t + 15
a) find the vertex by the method you are most comfortable with
in the form (t, h) , h will be your maximum height, t will be the time it happened.
b) clearly, when t = 0, h = 15 <---- the height of the window
c) set -5t^2 + 20t + 15 = 0 , and solve for t. Use the positive result of t
d) it will be the vertical line passing through your vertex, you found that in a)
To answer these questions, we need to analyze the given function.
The function h(t) = -5t^2 + 20t + 15 represents the height of the baseball at time t after it is thrown out of the window.
a) To find the maximum height that the baseball can reach, we need to determine the vertex of the parabola. The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula x = -b/2a.
In this case, a = -5 and b = 20. Using the formula, we can calculate the time at which the ball reaches its maximum height:
t = -20 / 2(-5) = 2 seconds
To find the height at this time, substitute t = 2 into the function h(t):
h(2) = -5(2)^2 + 20(2) + 15 = -20 + 40 + 15 = 35 meters
Therefore, the maximum height that the baseball can reach is 35 meters.
b) The height of the window is represented by the constant term in the function, which is 15 meters.
c) To find the time at which the ball hits the ground, we need to determine when the height h(t) becomes zero. Set h(t) = 0 and solve for t:
-5t^2 + 20t + 15 = 0
To simplify this equation, divide both sides by -5:
t^2 - 4t - 3 = 0
Now, we can factor or use the quadratic formula to solve for t. Factoring gives:
(t - 3)(t + 1) = 0
This equation has two solutions: t = 3 and t = -1. However, since time cannot be negative in this context, the ball hits the ground at t = 3 seconds.
d) The axis of symmetry of a parabola is a vertical line passing through the vertex. In this case, the vertex is located at t = 2 seconds (as found in part a). Therefore, the axis of symmetry is t = 2.
In summary:
a) The maximum height the baseball can reach is 35 meters.
b) The height of the window is 15 meters.
c) The ball hits the ground at 3 seconds.
d) The axis of symmetry of the function is t = 2.
a) To find the maximum height of the baseball, we need to determine the vertex of the parabolic function. The vertex of a parabolic function in the form of h(t) = at^2 + bt + c is given by the coordinates (-b/2a, f(-b/2a)). In our case, the equation is h(t) = -5t^2 + 20t + 15.
The coefficient of t^2 is a = -5, and the coefficient of t is b = 20. To find the x-coordinate of the vertex, we use the formula -b/2a:
t = -20 / (2 * -5) = -20 / -10 = 2
Now, substitute t = 2 into the equation h(t) to find the maximum height:
h(2) = -5(2)^2 + 20(2) + 15 = -20 + 40 + 15 = 35
Therefore, the maximum height the baseball can reach is 35 meters.
b) The height of the window can be determined by substituting t = 0 into the equation h(t):
h(0) = -5(0)^2 + 20(0) + 15 = 0 + 0 + 15 = 15
Thus, the window is at a height of 15 meters.
c) To find the time when the ball hits the ground, we set h(t) = 0 and solve for t:
-5t^2 + 20t + 15 = 0
We can simplify this equation by dividing all terms by -5:
t^2 - 4t - 3 = 0
Factoring the quadratic equation, we find:
(t - 3)(t + 1) = 0
Setting each factor equal to zero and solving for t gives us:
t - 3 = 0 ==> t = 3
t + 1 = 0 ==> t = -1
Since time cannot be negative in this context, the ball will hit the ground at t = 3 seconds.
d) The axis of symmetry of the parabolic function can be found using the formula -b/2a. In this case, a = -5 and b = 20. So:
t = -20 / (2 * -5) = 20 / 10 = 2
Therefore, the axis of symmetry of this function is at t = 2 seconds.