The stone should be thrown with minimum speed 24 feet/s so as to reach a height of 9 feet
Further explanation
To solve this problem there are several basic principles in Derivatives that need to be recalled, namely:
\(y = a ~ x^n \Rightarrow \frac{dy}{dx} = a ~ n ~ x^{n-1}\)
\(y = \sin x \Rightarrow \frac{dy}{dx} = \cos x\)
\(y = \cos x \Rightarrow \frac{dy}{dx} = - \sin x\)
\(y = u \times v \Rightarrow \frac{dy}{dx} = u' \times v + u \times v'\)
\(y = u \div v \Rightarrow \frac{dy}{dx} = \frac{u' \times v - u \times v'}{v^2}\)
\(y = u^n \Rightarrow \frac{dy}{dx} = n \times u^{n-1} \times u'\)
\(\text{where u and v are functions in variable x}
\)
\(\text{and u 'and v' are derivatives of u and v}\)
Let us now tackle the problem !
Given:
\(h(t) = -16t^2 + v_ot + h_o\)
The stone is thrown upward from ground level → h o = 0
\(h(t) = -16t^2 + v_ot + 0\)
\(h(t) = -16t^2 + v_ot\)
We will differentiate above equation to get the instantenous velocity of the stone.
\(v(t) = \frac{dh(t)}{dt}\)
\(v(t) = \frac{d}{dt} (-16t^2 + v_ot )\)
\(v(t) = -16(2t^{2-1}) + v_o\)
\(v(t) = -32t + v_o\)
At maximum height → v(t) = 0
\(v(t) = -32t + v_o\)
\(0 = -32t + v_o\)
\(32t = v_o\)
\(t = \boxed {\frac{v_o}{32}}\) → Equation 1
Let the maximum height h(t) = 9 feet , then :
\(h(t) = -16t^2 + v_ot\)
\(9 = -16( \frac{v_o}{32} )^2 + v_o( \frac{v_o}{32} )\) ← Equation 1
\(9 = -\frac{v_o^2}{64} + \frac{v_o^2}{32}\)
\(9 = \frac{v_o^2}{64}\)
\(v_o^2 = 9 \times 64\)
\(v_o = \sqrt{576}\)
\(v_o = \boxed {24 ~ feet/s}\)
Learn more
- Implicit Differentiation : https://brainly.com/question/4711711
- Logarithmic Differentiation : https://brainly.com/question/9226310
- Calculus Problem : https://brainly.com/question/11237537
Answer details
Grade: High School
Subject: Mathematics
Chapter: Differentiation
Keywords: Maximum , Minimum , Value , Function , Variable , Derivation , Differentiation
