If the ball was thrown straight up at 24 ft/sec when it was 5 ft above the ground, the ball reached a maximum height of 7.25 m
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Discriminant of quadratic equation ( ax² + bx + c = 0 ) could be calculated by using :
From the value of Discriminant , we know how many solutions the equation has by condition :
D < 0 → No Real Roots
D = 0 → One Real Root
D > 0 → Two Real Roots
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An axis of symmetry of quadratic equation y = ax² + bx + c is :
[tex]\large {\boxed {x = \frac{-b}{2a} } }[/tex]
Let us now tackle the problem!
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Given:
[tex]p(t) = 3gt^2 + v_ot + p_o[/tex]
[tex]p(t) = 3(-32)t^2 + 24t + 5[/tex]
[tex]p(t) = -64t^2 + 24t + 5[/tex]
Asked:
[tex]p_{max} = ?[/tex]
Solution:
At the maximum height , velocity is 0 m/s:
[tex]v = \frac{dp(t)}{dt}[/tex]
[tex]v = \frac{d}{dt} ( -64t^2 + 24t + 5 )[/tex]
[tex]v = (-64)(2)t^{2-1} + 24[/tex]
[tex]v = -128t + 24[/tex]
[tex]0 = -128t + 24[/tex]
[tex]128t = 24[/tex]
[tex]t = 24 \div 128[/tex]
[tex]t = 3 \div 16[/tex]
[tex]t = 0.1875 \texttt{ s}[/tex]
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[tex]p(t) = -64t^2 + 24t + 5[/tex]
[tex]p(0.1875) = -64(0.1875)^2 + 24(0.1875) + 5[/tex]
[tex]p(0.1875) = 7.25 \texttt{ m}[/tex]
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Grade: High School
Subject: Mathematics
Chapter: Quadratic Equations
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Keywords: Quadratic , Equation , Discriminant , Real , Number
#LearnWithBrainly
Step-by-step explanation:
Equation is given by p(t) = 3gt² + v₀t + P₀
Given that
g = -32 ft/s²
v₀ = 24 ft/s
P₀ = 5 ft
Substituting
p(t) = 3 x -32 x t² + 24 x t + 5
p(t) = -96 t² + 24 t + 5
We need to find maximum of this equation, at maximum we have derivative is zero.
p'(t) = -96 x 2t + 24 = 0
192 t = 24
t = 0.125 s
Substituting in p(t) equation
p(0.125) = -96 x 0.125² + 24 x 0.125 + 5
p(0.125) = 6.5 ft
Maximum height reached is 6.5 ft
