Suppose a ball is thrown straight up into the air, and the height of the ball above the ground is given by the function h(t) = 6 + 37t – 16t2, where h is in feet and t is in seconds. At what time t does the ball stop going up and start returning to earth?
37 - 32(3.2) = -65.4
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To find the time at which the ball stops going up and starts returning to earth, we need to find the vertex of the parabolic function h(t) = 6 + 37t – 16t^2. The vertex of a parabola represents the highest or lowest point on the graph.
The equation of a parabola in the form h(t) = a + bt + ct^2 has a vertex at t = -b/2c. In this case, a = 6, b = 37, and c = -16.
Substituting these values into the formula, we get:
t = -(37) / (2 * (-16))
Simplifying the expression further:
t = 37/32
Therefore, the ball stops going up and starts returning to earth at t = 37/32 seconds.
you are given h(t), to you need to take the derivatave of h'(t) and set it to zero.
The derivative of height is velocity.