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.
shift coursejohn
To find the maximum height, we need to determine the highest point on the path of the ball. This occurs when the velocity of the ball becomes zero.
First, find the velocity of the ball by taking the derivative of the height equation with respect to time (t):
v = ds/dt = d/dt [(-1/27)(t^2) + 4√t]
Taking the derivative term by term:
v = (-1/27)(2t) + 4(1/2)(t^(-1/2))
Simplifying:
v = (-2/27)t + 2√t
Next, set the velocity equation equal to zero to find when the ball reaches its highest point:
0 = (-2/27)t + 2√t
Rearranging the equation:
(2/27)t = 2√t
Squaring both sides:
(4/729)t^2 = 4t
Simplifying:
(1/729)t^2 - t = 0
Factoring out t:
t((1/729)t - 1) = 0
This equation has two solutions: t = 0 and t = 729. However, the ball is thrown vertically into the air, so t = 0 is not relevant.
Thus, the maximum height occurs at t = 729 seconds.
To find the maximum height, substitute 729 into the height equation:
s = (-1/27)(729^2) + 4√729
Simplifying:
s = (-1/27)(531441) + 36
s ≈ -19683 + 36
s ≈ -19647
Therefore, the maximum height of the ball is approximately -19647 units.