By applying the <em>quadratic</em> formula and discriminant of the <em>quadratic</em> formula, we find that the <em>maximum</em> height of the ball is equal to 75.926 meters.
<h3>How to determine the maximum height of the ball</h3>
Herein we have a <em>quadratic</em> equation that models the height of a ball in time and the <em>maximum</em> height represents the vertex of the parabola, hence we must use the <em>quadratic</em> formula for the following expression:
- 4.8 · t² + 19.9 · t + (55.3 - h) = 0
The height of the ball is a maximum when the discriminant is equal to zero:
19.9² - 4 · (- 4.8) · (55.3 - h) = 0
396.01 + 19.2 · (55.3 - h) = 0
19.2 · (55.3 - h) = -396.01
55.3 - h = -20.626
h = 55.3 + 20.626
h = 75.926 m
By applying the <em>quadratic</em> formula and discriminant of the <em>quadratic</em> formula, we find that the <em>maximum</em> height of the ball is equal to 75.926 meters.
To learn more on quadratic equations: brainly.com/question/17177510
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Answer:
x=6
Step-by-step explanation:
Answer:
Step-by-step explanation:
height h = 11 cm
slant height s = 11.180339887499 cm
side length a = 4 cm
lateral edge length e = 11.357816691601 cm
1/2 side length r = 2 cm
volume V = 58.666666666667 cm3
lateral surface area L = 89.442719099992 cm2
base surface area B = 16 cm2
total surface area A = 105.44271909999 cm2
C . the graph will shift upwards because it is increasing on a graph
If you've started pre-calculus, then you know that the derivative of h(t)
is zero where h(t) is maximum.
The derivative is h'(t) = -32 t + 96 .
At the maximum ... h'(t) = 0
32 t = 96 sec
t = 3 sec .
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If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.
In that case, the question GIVES you the maximum height.
Just write it in place of h(t), then solve the quadratic equation
and find out what 't' must be at that height.
150 ft = -16 t² + 96 t + 6
Subtract 150ft from each side: -16t² + 96t - 144 = 0 .
Before you attack that, you can divide each side by -16,
making it a lot easier to handle:
t² - 6t + 9 = 0
I'm sure you can run with that equation now and solve it.
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.
(Funny how the two widely different methods lead to the same answer.)
The answer is from AL2006
