A ball is thrown vertically into the air so that its height s after t seconds is given by the equation, s=-1/27(t^2)+4√t. Find its maximum height.
ds/dt = -2/27 t + 2/√t
= (54-2t^(3/2))/(27√t)
ds/dt=0 when
54 - 2t^(3/2) = 0
t^(3/2) = 27
t = 9
now just find s(9)
Well, to find the maximum height, we need to find the vertex of the parabolic equation. Since we have a negative coefficient for the squared term, the parabola opens downward, meaning the vertex represents the highest point.
To find the vertex, we can use the formula -b/2a, where a is the coefficient of the squared term and b is the coefficient of the linear term.
In this case, a = -1/27 and b = 4√t. Plugging in these values, we get:
t = -4√t / (2 * (-1/27))
Simplifying further:
t = 6√t
Now, let's square both sides to eliminate the square root:
t^2 = (6√t)^2
t^2 = 36t
Rearranging the equation:
t^2 - 36t = 0
Factorizing:
t(t - 36) = 0
So the possible values for t are t = 0 and t = 36. However, since we're looking for the maximum height, we discard t = 0 as it represents the starting point.
Now, plug t = 36 back into the equation s = -1/27(t^2)+4√t:
s = -1/27(36^2) + 4√36
= -1/27(1296) + 4 * 6
= -48 + 24
= -24
So, the maximum height of the ball is -24 units. I must say, that's a pretty low ball! But hey, at least it's still a height, right? Keep your head up!
To find the maximum height of the ball, we need to find the vertex of the parabolic equation. The vertex can be found using the formula h = -b/2a, where the equation is in the form s = at^2 + bt + c.
In this case, the equation for the height s is given as s = -1/27(t^2) + 4√t. To put it in the required form, rewrite it as s = -1/27(t^2) + 4t^(1/2).
Comparing this to the general form s = at^2 + bt + c, we see that a = -1/27, b = 4, and c = 0.
Now, we can find the vertex by using the formula h = -b/2a:
h = -4 / (2 * (-1/27))
h = -4 / (-2/27)
h = -4 * (-27/2)
h = 54
Therefore, the maximum height of the ball is 54 units.
To find the maximum height of the ball, we need to find the vertex of the parabolic equation s = -(1/27)t^2 + 4√t. The vertex form of a parabolic equation is given by s = a(t - h)^2 + k, where (h, k) represents the coordinates of the vertex.
Comparing the given equation, s = -(1/27)t^2 + 4√t, with the vertex form, we can see that a = -(1/27).
To find the x-coordinate of the vertex (h), we use the formula h = -b / (2a), where b is the coefficient of t. In this case, b would be zero since there is no t term in the equation. Therefore, h = -0 / (2 * -(1/27)) = 0.
Next, we can find the y-coordinate of the vertex (k) by substituting the value of h into the equation. In this case, k = -(1/27) * (0)^2 + 4√0 = 0.
So, the coordinates of the vertex are (0, 0). The maximum height of the ball is 0 units.
This means that the ball does not reach any positive height and falls straight down from where it was thrown.