If a ball is thrown vertically upward from the roof of 48 foot building with a velocity of 32 ft/sec, its height after t seconds is s(t)=48+32t–16t2. What is the maximum height the ball reaches?
s'(t) = 32 - 32t
= 0 for a max/min
t = 1
then s(1) = 48 + 32 - 16 = 64
max height is 64 above the ground, or 16 feet above the roof line
To find the maximum height the ball reaches, we need to determine the vertex of the parabolic function s(t) = 48 + 32t - 16t^2.
The vertex of a parabola is given by the formula (h, k) where h represents the time at which the maximum height is reached, and k represents the maximum height.
The formula for h is: h = -b / (2a)
Where a, b, and c are the coefficients of the quadratic equation.
Given the equation s(t) = 48 + 32t - 16t^2, we can see that a = -16 and b = 32.
Therefore, h = -(32) / (2*(-16))
= -32 / (-32)
= 1
So, the ball reaches its maximum height at t = 1 second.
To find k, we substitute the value of t into the equation s(t):
s(1) = 48 + 32(1) - 16(1)^2
= 48 + 32 - 16
= 80 - 16
= 64
Therefore, the maximum height the ball reaches is 64 feet.
To find the maximum height the ball reaches, we need to determine the vertex of the parabolic function s(t) = 48 + 32t - 16t^2. The vertex of a parabola is the maximum or minimum point of the graph.
The equation for the vertex of a parabola in the form of s(t) = a(t - h)^2 + k is (h, k). In our case, a = -16, which means the parabola opens downward.
To find the vertex, we need to use the formula:
h = -b / (2a)
In our equation, b = 32 and a = -16, so:
h = -32 / (2 * -16)
= -32 / -32
= 1
Substituting h = 1 back into the equation for s(t), we can find the maximum height of the ball:
s(1) = 48 + 32(1) - 16(1^2)
= 48 + 32 - 16
= 64
Therefore, the maximum height the ball reaches is 64 feet.