If you are standing near the edge of the top of a 200 feet building and throw a ball vertically upward it will be modeled by the function s(t)=-16t^2+64t+200 where s(t) is the ball's height above ground in feet and t is seconds after the ball was thrown.
A) When does the ball reach its maximum height and what is the maximum height?
T= -b/2a
T= -64/-2(-16)
T= 2 seconds
How do I find the maximum height?
B) When does the ball hit the ground? (Round to the nearest tenth of a second)
Can you please help me with this question?
C) Find s(0) and describe what it represents?
Can you please help me with this question?
D) How would I graph the quadratic function
Please help me solving each parts step by step!!!!
A. Plug your calculated value of T(2 s.)
into the given Eq and solve for S, the ht.
B. Tr = 2 s. = Rise time(part A).
h = 0.5g*Tf^2 = 264 Ft.
g = 32Ft/s^2.
Tf = Fall time.
Tf = ?
Tr + Tf = Time to reach gnd.
C. S(o) = ht. when t = 0. In the given Eq, replace t with zero and solve for S(o).
D. Using the given Eq, calculate the ht.
for each value of T. Use the data for graphing.
(T, h).
(0.0,200).
(0.5, ).
(1.0,248).
(1.5, ).
V(2.0,264).
(2.5, ).
(3.0,248).
(3.5, ).
(4.0,200).
A) To find the maximum height of the ball, you can use the formula T = -b/2a, where T represents the time at which the ball reaches its maximum height. In your case, a = -16 (the coefficient of the t^2 term) and b = 64 (the coefficient of the t term).
Plugging these values into the formula, we have:
T = -b/2a = -64 / (2 * (-16)) = 2 seconds
So, the ball reaches its maximum height after 2 seconds.
To find the maximum height, you can substitute the value of T (= 2 seconds) into the equation s(t) = -16t^2 + 64t + 200 and calculate s(2):
s(2) = -16(2)^2 + 64(2) + 200 = -64 + 128 + 200 = 264 feet
Therefore, the maximum height of the ball is 264 feet.
B) To determine when the ball hits the ground, you need to find the value of t when s(t) = 0. Since the equation represents the height of the ball, when it hits the ground, the height is zero.
So, we can rewrite the equation as: -16t^2 + 64t + 200 = 0
To solve this quadratic equation, you can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 64, and c = 200. Plugging these values into the quadratic formula, we get:
t = (-64 ± √(64^2 - 4 * (-16) * 200)) / (2 * (-16))
After simplifying the equation further, you would find two solutions for t. However, we are only interested in the positive value of t since we are dealing with time.
Rounding to the nearest tenth of a second, t ≈ 5.8 seconds.
Therefore, the ball hits the ground approximately 5.8 seconds after being thrown.
C) To find s(0), you need to substitute t = 0 into the equation s(t) = -16t^2 + 64t + 200:
s(0) = -16(0)^2 + 64(0) + 200 = 200
s(0) represents the initial height of the ball above the ground. In this case, s(0) = 200, meaning that when the ball is released (t = 0), it is already 200 feet above the ground.
D) To graph the quadratic function s(t) = -16t^2 + 64t + 200, you can follow these steps:
1. Determine the x-axis and y-axis range based on the given problem. In this case, the x-axis represents time (t) and the y-axis represents the height (s(t)).
2. Choose specific values for t and calculate the corresponding values for s(t). Plot these points on the graph.
3. Since this is a quadratic function, the graph will be a downward-facing parabola. Additionally, as the coefficient of the t^2 term is negative (-16), the parabola will open downwards.
4. Use the vertex formula, T = -b/2a, to find the x-coordinate of the vertex. In this case, the vertex represents the maximum height.
5. Plot the vertex on the graph, which will be the highest point on the parabola.
6. Connect the points and draw the parabolic curve, taking into consideration that the shape will gradually descend below the x-axis as t values increase.
Remember to label the x-axis as "time (t)" and the y-axis as "height (s)."