The height in feet of a projectile with an initial velocity of 32 feet per second and an initial height of 48 feet is a function of time in seconds given by
h(t) = −16t2 + 32t + 48. (a) Find the maximum height of the projectile.
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
complete the square to find the vertex if you do not know calculus
16 t^2 -32 t -48 = -h
t^2 - 2 t - 3 = -h/16
t^2 - 2 t = -h/16 + 3
t^2 - 2 t + 1 = -h/16 + 4 = -1/16(h-64)
(t-1)^2 = -(1/16) (h-64)
max height is vertex at t = 1 second and h = 64 ft
c) t^2 - 2 t - 3 = 0
(t-3)(t+1) = 0
t = 3 when h = 0
(a) The maximum height of the projectile can be found by determining the vertex of the parabolic function. The vertex can be found using the formula t = -b/2a, where a = -16 and b = 32.
t = -32 / (2*(-16))
t = 32 / 32
t = 1
Substituting t = 1 into the function, we can find the maximum height:
h(1) = -16(1)^2 + 32(1) + 48
h(1) = -16 + 32 + 48
h(1) = 64 ft
Therefore, the maximum height of the projectile is 64 ft.
(b) The time when the projectile achieves its maximum height can be found from the vertex. As calculated in part (a), the time is t = 1 second.
Therefore, the projectile achieves its maximum height at t = 1 second.
(c) To find the time when the projectile has a height of 0 feet, we need to set the function h(t) equal to 0 and solve for t:
0 = -16t^2 + 32t + 48
Using the quadratic formula:
t = (-32 ± √(32^2 - 4(-16)(48)) / (2(-16))
t = (-32 ± √(1024 + 3072) / -32
t = (-32 ± √(4096) / -32
t = (-32 ± 64) / -32
t = (-32 + 64) / -32 or t = (-32 - 64) / -32
t = 32 / -32 or t = 96 / -32
t = -1 or t = -3
Therefore, the projectile has a height of 0 feet at t = -1 second or t = -3 seconds.
(a) To find the maximum height of the projectile, we need to find the vertex of the quadratic function h(t) = -16t^2 + 32t + 48. The vertex can be found using the formula t = -b / (2a), where a = -16 and b = 32.
Substituting the values into the formula, we have:
t = -32 / (2 * (-16))
t = -32 / (-32)
t = 1
To find the maximum height, substitute this value of t into the function h(t):
h(1) = -16(1)^2 + 32(1) + 48
h(1) = -16 + 32 + 48
h(1) = 64 ft
Therefore, the maximum height of the projectile is 64 ft.
(b) The time when the projectile achieves its maximum height is t = 1 second.
(c) To find the time when the projectile has a height of 0 feet, we need to set h(t) equal to 0 and solve for t:
0 = -16t^2 + 32t + 48
We can solve this equation by factoring or using the quadratic formula. Factoring gives:
0 = 16(t^2 - 2t - 3)
0 = 16(t - 3)(t + 1)
Setting each factor equal to zero, we get:
t - 3 = 0 or t + 1 = 0
t = 3 or t = -1
Since time cannot be negative in this context, the projectile reaches a height of 0 ft at t = 3 seconds.
Therefore, the time when the projectile has a height of 0 feet is t = 3 seconds.
(a) To find the maximum height of the projectile, we need to determine the vertex of the quadratic function h(t) = -16t^2 + 32t + 48. The vertex of a quadratic function in the form of h(t) = ax^2 + bx + c can be found using the formula x = -b/2a.
In this case, a = -16 and b = 32. Substituting these values into the formula, we get:
t = -32 / (2 * -16) = -32 / -32 = 1
To find the maximum height, we substitute the value of t back into the equation:
h(t) = -16(1)^2 + 32(1) + 48 = -16 + 32 + 48 = 64
Therefore, the maximum height of the projectile is 64 feet.
(b) The time t when the projectile achieves its maximum height is the x-coordinate of the vertex. In this case, we found earlier that t = 1. Therefore, the projectile achieves its maximum height at t = 1 second.
(c) To find the time t when the projectile has a height of 0 feet, we set h(t) = 0 and solve for t. Using the quadratic equation formula, we can find the solutions for t.
h(t) = -16t^2 + 32t + 48
Setting h(t) = 0:
-16t^2 + 32t + 48 = 0
We can solve this equation by factoring, completing the square, or using the quadratic formula. In this case, using the quadratic formula will give us the solutions:
t = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values in the equation above:
t = (-32 ± √(32^2 - 4(-16)(48))) / (2(-16))
t = (-32 ± √(1024 + 3072)) / (-32)
t = (-32 ± √4096) / (-32)
t = (-32 ± 64) / (-32)
Simplifying this expression, we get:
t = (-32 + 64) / (-32) or t = (-32 - 64) / (-32)
t = 32 / -32 or t = -96 / -32
t = -1 or t = 3
Therefore, the projectile has a height of 0 feet at t = -1 second and t = 3 seconds.