Maximum height of the ball is the highest values in the range of values of
the height of the ball.
The time it takes the ball to reach the maximum height is 1.5 seconds.
Reasons:
The given function for the height of the ball is h(t) = -16·t² + 48·t + 100
The upward velocity of the ball, v = 48 ft./s
Height of the platform from which the ball is launched = 100 ft.
Required:
The time it takes the ball to reach maximum height.
Solution:
The function is a quadratic function, with a negative leading coefficient,
and therefore, has a maximum point, which is given by the value of t at the
point
where; [tex]\dfrac{d(h(t))}{dt} = 0[/tex]
Therefore, we have;
[tex]\dfrac{d(h(t))}{dt} = \dfrac{d(-16 \cdot t^2 + 48 \cdot t + 100)}{dt} = -32 \cdot t + 48 = 0[/tex]
[tex]\mathrm{At \ the \ maximum \ height, } \ t = \dfrac{48}{32} = 1.5[/tex]
The time it takes the ball to reach the maximum height, t = 1.5 seconds
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