The stone should be thrown with minimum speed 24 feet/s so as to reach a height of 9 feet
To solve this problem there are several basic principles in Derivatives that need to be recalled, namely:
[tex]y = a ~ x^n \Rightarrow \frac{dy}{dx} = a ~ n ~ x^{n-1}[/tex]
[tex]y = \sin x \Rightarrow \frac{dy}{dx} = \cos x[/tex]
[tex]y = \cos x \Rightarrow \frac{dy}{dx} = - \sin x[/tex]
[tex]y = u \times v \Rightarrow \frac{dy}{dx} = u' \times v + u \times v'[/tex]
[tex]y = u \div v \Rightarrow \frac{dy}{dx} = \frac{u' \times v - u \times v'}{v^2}[/tex]
[tex]y = u^n \Rightarrow \frac{dy}{dx} = n \times u^{n-1} \times u'[/tex]
[tex]\text{where u and v are functions in variable x}
[/tex]
[tex]\text{and u 'and v' are derivatives of u and v}[/tex]
Let us now tackle the problem !
Given:
[tex]h(t) = -16t^2 + v_ot + h_o[/tex]
The stone is thrown upward from ground level → h o = 0
[tex]h(t) = -16t^2 + v_ot + 0[/tex]
[tex]h(t) = -16t^2 + v_ot[/tex]
We will differentiate above equation to get the instantenous velocity of the stone.
[tex]v(t) = \frac{dh(t)}{dt}[/tex]
[tex]v(t) = \frac{d}{dt} (-16t^2 + v_ot )[/tex]
[tex]v(t) = -16(2t^{2-1}) + v_o[/tex]
[tex]v(t) = -32t + v_o[/tex]
At maximum height → v(t) = 0
[tex]v(t) = -32t + v_o[/tex]
[tex]0 = -32t + v_o[/tex]
[tex]32t = v_o[/tex]
[tex]t = \boxed {\frac{v_o}{32}}[/tex] → Equation 1
Let the maximum height h(t) = 9 feet , then :
[tex]h(t) = -16t^2 + v_ot[/tex]
[tex]9 = -16( \frac{v_o}{32} )^2 + v_o( \frac{v_o}{32} )[/tex] ← Equation 1
[tex]9 = -\frac{v_o^2}{64} + \frac{v_o^2}{32}[/tex]
[tex]9 = \frac{v_o^2}{64}[/tex]
[tex]v_o^2 = 9 \times 64[/tex]
[tex]v_o = \sqrt{576}[/tex]
[tex]v_o = \boxed {24 ~ feet/s}[/tex]
Grade: High School
Subject: Mathematics
Chapter: Differentiation
Keywords: Maximum , Minimum , Value , Function , Variable , Derivation , Differentiation
The minimum speed to reach a height of 9 feet is 24 feet per second
The height function is given as:
[tex]h(t) = -16t^2 +vt[/tex]
Differentiate the above function
[tex]h'(t) = -32t +v[/tex]
Set to 0
[tex]-32t +v = 0[/tex]
Subtract v from both sides
[tex]-32t = -v[/tex]
Divide both sides by -32
[tex]t = \frac{v}{32}[/tex]
The maximum height is given as: 9 ft.
At this point, the value of t is [tex]\frac{v}{32}[/tex]
i.e.
[tex]h(\frac{v}{32}) = 9[/tex]
So, we have:
[tex]9 = -16(\frac{v}{32})^2 +(\frac{v}{32})v[/tex]
Expand
[tex]9 = -16(\frac{v^2}{1024}) +(\frac{v}{32})v[/tex]
[tex]9 = -\frac{v^2}{64} +(\frac{v}{32})v[/tex]
Expand the brackets
[tex]9 = -\frac{v^2}{64} +\frac{v^2}{32}[/tex]
Take LCM
[tex]9 = \frac{-v^2 + 2v^2}{64}[/tex]
[tex]9 = \frac{v^2}{64}[/tex]
Multiply both sides by 64
[tex]v^2= 64 \times9[/tex]
Take square roots of both sides
[tex]v= 8 \times3[/tex]
[tex]v= 24[/tex]
Hence, the minimum speed to reach a height of 9 feet is 24 feet per second
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