ritvik tosses a ball from where he stands on a balcony to his friend on the ground. the height, in feet, of the ball can be represented by the function d(t) = 16t to the power of 2 + 16t + 9 where t represents time in seconds. part a: from what height was the ball tossed? part b: what is the maximum height of the ball. part c: how long does it take the ball to reach its maximum height?
To find the answers to these questions, we need to analyze the given function and its properties. Let's break down each part separately.
Part a: From what height was the ball tossed?
To determine the initial height, we need to find the value of the function when t = 0. Simply substitute t = 0 into the function d(t):
d(0) = 16(0)^2 + 16(0) + 9
= 0 + 0 + 9
= 9
Therefore, the ball was initially tossed from a height of 9 feet.
Part b: What is the maximum height of the ball?
The maximum height of the ball will occur at the vertex of the quadratic function. The vertex of a quadratic function in the form ax^2 + bx + c can be found using the formula:
x = -b / (2a)
In this case, the function d(t) = 16t^2 + 16t + 9 has a = 16 and b = 16.
t = -16 / (2 * 16)
= -16 / 32
= -0.5
To find the maximum height, substitute t = -0.5 into d(t):
d(-0.5) = 16(-0.5)^2 + 16(-0.5) + 9
= 16(0.25) - 8 + 9
= 4 - 8 + 9
= 5
Therefore, the maximum height reached by the ball is 5 feet.
Part c: How long does it take the ball to reach its maximum height?
We already found that the t-coordinate of the vertex is -0.5. Therefore, it takes t = -0.5 seconds for the ball to reach its maximum height.
Note: Since time cannot be negative in this context, it can be interpreted that it takes 0.5 seconds (half a second) for the ball to reach its maximum height.
d(t) = -16t^2 + 16t + 9
(a) d(0) = 9
(c) the vertex is at t = -b/2a = 1/2.
(b) find d(1/2)