The height in feet of a projectile with an initial velocity of 96 feet per second and an initial height of 112 feet is a function of time in seconds given by
h(t) = −16t2 + 96t + 112.
(a) Find the maximum height of the projectile.
144
Incorrect: Your answer is incorrect.
ft
(b) Find the time t when the projectile achieves its maximum height.
t =
sec
(c) Find the time t when the projectile has a height of 0 feet.
t =
sec
the max height is where t = -b/2a = 3
h(3) = 256 ft
h=0 when t=7
(a) To find the maximum height of the projectile, we need to find the vertex of the quadratic function h(t) = −16t^2 + 96t + 112. The vertex can be found using the formula x = -b / (2a) where a, b, and c are the coefficients of the quadratic equation.
In this case, a = -16 and b = 96. Plugging these values into the formula, we get:
t = -96 / (2 * -16)
t = -96 / -32
t = 3
Now we can find the height by plugging this value of t back into the equation:
h(t) = −16(3)^2 + 96(3) + 112
h(t) = -16 * 9 + 96 * 3 + 112
h(t) = -144 + 288 + 112
h(t) = 256
Therefore, the maximum height of the projectile is 256 feet.
(b) To find the time when the projectile achieves its maximum height, we already found it in part (a). The time t is 3 seconds.
(c) To find the time when the projectile has a height of 0 feet, we need to set the equation h(t) = −16t^2 + 96t + 112 to 0 and solve for t.
-16t^2 + 96t + 112 = 0
We can solve this equation by factoring or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 96, and c = 112. Plugging these values into the formula, we get:
t = (-96 ± √(96^2 - 4 * -16 * 112)) / (2 * -16)
t = (-96 ± √(9216 + 7168)) / -32
t = (-96 ± √(16384)) / -32
t = (-96 ± 128) / -32
Now we have two possible solutions:
t1 = (-96 + 128) / -32
t1 = 32 / -32
t1 = -1
t2 = (-96 - 128) / -32
t2 = -224 / -32
t2 = 7
Therefore, the projectile reaches a height of 0 feet at t = -1 second and t = 7 seconds.
To find the maximum height of the projectile, we need to determine the vertex of the quadratic function h(t) = -16t^2 + 96t + 112.
(a) The maximum height can be found by finding the y-coordinate of the vertex. The formula for the x-coordinate of the vertex is given by x = -b / (2a), where a is the coefficient of the t^2 term (-16 in this case) and b is the coefficient of the t term (96 in this case).
Using the formula, we have:
x = -96 / (2 * (-16))
x = -96 / (-32)
x = 3
To find the y-coordinate, substitute the value of t = 3 into the quadratic function:
h(3) = -16(3)^2 + 96(3) + 112
h(3) = -144 + 288 + 112
h(3) = 256
Thus, the maximum height of the projectile is 256 feet.
(b) To find the time when the projectile achieves its maximum height, we already know that the x-coordinate of the vertex is 3. Therefore, the time t when the projectile achieves its maximum height is t = 3 seconds.
(c) To find the time when the projectile has a height of 0 feet, we need to solve the quadratic equation h(t) = -16t^2 + 96t + 112 = 0. This can be done by factoring, completing the square, or using the quadratic formula.
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients in the quadratic equation.
In this case, a = -16, b = 96, and c = 112. Plugging these values into the quadratic formula, we have:
t = (-96 ± √(96^2 - 4(-16)(112))) / (2 * (-16))
t = (-96 ± √(9216 + 7168)) / (-32)
t = (-96 ± √16384) / (-32)
t = (-96 ± 128) / (-32)
Simplifying further, we get two possible solutions:
t = (-96 + 128) / (-32) = 32 / (-32) = -1
t = (-96 - 128) / (-32) = -224 / (-32) = 7
Since time cannot be negative in this context, the time t when the projectile has a height of 0 feet is t = 7 seconds.