A baseball is throw from the top of a building and falls to the ground below. Its path is approximated by the relation h= -5t^2 +5t +30, where h is the height above ground in meters and t is the elapsed time in seconds.
a. How tall is the building/
b. when does the ball reach its maximum height?
c. How high above the building is the ball at its maximum height?
Where do I start?
Well, to start, I would suggest finding the value of h when t = 0. This will give you the initial height of the ball, which corresponds to the height of the building. So let's put t = 0 into the equation:
h = -5(0)^2 + 5(0) + 30
Simplifying that, we get:
h = 30
So the building is 30 meters tall.
Next, let's find the time when the ball reaches its maximum height. To do that, we can take the derivative of h with respect to t and set it equal to zero. That's where the maximum point should be. So let's differentiate:
dh/dt = -10t + 5
Setting this equal to zero gives us:
-10t + 5 = 0
Solving for t, we get:
t = 0.5
So the ball reaches its maximum height at 0.5 seconds.
Finally, to find how high above the building the ball is at its maximum height, we can plug t = 0.5 back into the equation for h:
h = -5(0.5)^2 + 5(0.5) + 30
Calculating that gives us:
h = 32.5
So at its maximum height, the ball is 2.5 meters above the building.
I hope that helps! Let me know if you have any more questions.
To answer these questions, we need to understand the given relation and its properties. The relation h = -5t^2 + 5t + 30 represents the height of the ball above the ground at a given time, where h is the height in meters and t is the elapsed time in seconds.
a. How tall is the building?
To find the height of the building, we need to determine the value of h when t = 0. Substituting t = 0 into the equation, we get h = -5(0)^2 + 5(0) + 30 = 0 + 0 + 30 = 30. Therefore, the building is 30 meters tall.
b. When does the ball reach its maximum height?
To find the time when the ball reaches its maximum height, we can determine the vertex of the parabolic equation h = -5t^2 + 5t + 30. The vertex occurs at the value of t that minimizes or maximizes the function. The formula for the t-coordinate of the vertex of a parabolic equation is given by t = -b / (2a), where a and b are coefficients of t^2 and t respectively. In this case, a = -5 and b = 5. Therefore, t = -5 / (2(-5)) = -5 / -10 = 0.5 seconds. So, the ball reaches its maximum height at t = 0.5 seconds.
c. How high above the ground is the ball at its maximum height?
To find the height of the ball at its maximum height, we substitute the value of t = 0.5 into the equation h = -5t^2 + 5t + 30.
h = -5(0.5)^2 + 5(0.5) + 30
= -5(0.25) + 2.5 + 30
= -1.25 + 2.5 + 30
= 1.25 + 30
= 31.25
So, the ball is 31.25 meters above the ground at its maximum height.
To answer these questions, we need to understand the given equation and its properties. The equation provided for the height of the ball above the ground is given by h = -5t^2 + 5t + 30.
(a) To find the height of the building, we need to determine the value of h when t = 0. When t = 0, the term -5t^2 becomes 0, leaving us with h = 5 * 0 + 30 = 30 meters. Therefore, the height of the building is 30 meters.
(b) To find the time at which the ball reaches its maximum height, we need to determine the vertex of the parabolic equation, which represents the highest point of the ball's path. The equation is in the form h = at^2 + bt + c, where a = -5, b = 5, and c = 30.
The formula for the x-coordinate of the vertex of a quadratic equation in the form ax^2 + bx + c is given by x = -b/2a. In this case, we substitute b = 5 and a = -5 into the formula to find t = -5/(2*(-5)) = 0.5 seconds. Therefore, the ball reaches its maximum height at 0.5 seconds.
(c) To determine the height of the ball at its maximum height, we substitute the value of t = 0.5 into the original equation. h = -5*(0.5)^2 + 5*(0.5) + 30 = -1.25 + 2.5 + 30 = 31.25 meters. Therefore, the ball is 31.25 meters above the ground at its maximum height.
To summarize:
(a) The height of the building is 30 meters.
(b) The ball reaches its maximum height at 0.5 seconds.
(c) The ball is 31.25 meters above the building at its maximum height.
You must be studying parabolas to have this type of question.
You must have learned how to find the vertex of a parabola
find the vertex (t, h)
the t is the time when you reach the maximum height
the h is the maximum height obtained.
as to the height of the building, let t = 0