Find the distance between the points (-7, -10) and (-5, -1)
1 answer:
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We assume you intend h(t) to be the usual equation of ballistic motion
h(t) = -16t² + 40t + 10
a. The function can be written in vertex form as
h(t) = -16(t² -5/2t) +10
h(t) = -16(t² -5/2t +(5/4)²) + 10 +5²
h(t) = -16(t -5/4)² + 35
This is the vertex form a(x-h)²+k where (h, k) is the vertex of the parabola. The maximum height will be "k" for a < 0, which it is.
The ball's maximum height is 35 ft.
b. h(t) = 0 for the value of t that satisfies
0 = -16(t -5/4)² + 35
(t -5/4)² = 35/16
t = (5 +√35)/4 ≈ 2.729
About 2.729 seconds have passed when the ball hits the ground.
_____
I find it convenient to let a graphing calculator tell you the answers to these questions. As above, the maximum height is the y-coordinate of the vertex. The elapsed time is the x-coordinate of the positive x-intercept.
h(t) = -16t² + 40t + 10
a. The function can be written in vertex form as
h(t) = -16(t² -5/2t) +10
h(t) = -16(t² -5/2t +(5/4)²) + 10 +5²
h(t) = -16(t -5/4)² + 35
This is the vertex form a(x-h)²+k where (h, k) is the vertex of the parabola. The maximum height will be "k" for a < 0, which it is.
The ball's maximum height is 35 ft.
b. h(t) = 0 for the value of t that satisfies
0 = -16(t -5/4)² + 35
(t -5/4)² = 35/16
t = (5 +√35)/4 ≈ 2.729
About 2.729 seconds have passed when the ball hits the ground.
_____
I find it convenient to let a graphing calculator tell you the answers to these questions. As above, the maximum height is the y-coordinate of the vertex. The elapsed time is the x-coordinate of the positive x-intercept.