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Conservation of energy A projectile with mass m is
Chapter 11, Problem 53E(choose chapter or problem)
Conservation of energy A projectile is launched into the air on a parabolic trajectory. For t \(\geq\) 0, its horizontal and vertical coordinates are \(x(t)=u_{0} t \text { and } y(t)=-(1 / 2) g t^{2}+v_{0} t\), respectively, where \(u_{0}\) is the initial horizontal velocity, \(v_{0}\) is the initial vertical velocity, and g is the acceleration due to gravity. Recalling that u(t) = x'(t) and v(t) = y’(t) are the components of the velocity, the energy of the projectile (kinetic plus potential) is
\(E(t)=\frac{1}{2} m\left(u^{2}+v^{2}\right)+m g y\)
Use the Chain Rule to compute E'(t) and show that E' (t) = 0 for all t \(\geq\) 20. Interpret the result.
Questions & Answers
QUESTION:
Conservation of energy A projectile is launched into the air on a parabolic trajectory. For t \(\geq\) 0, its horizontal and vertical coordinates are \(x(t)=u_{0} t \text { and } y(t)=-(1 / 2) g t^{2}+v_{0} t\), respectively, where \(u_{0}\) is the initial horizontal velocity, \(v_{0}\) is the initial vertical velocity, and g is the acceleration due to gravity. Recalling that u(t) = x'(t) and v(t) = y’(t) are the components of the velocity, the energy of the projectile (kinetic plus potential) is
\(E(t)=\frac{1}{2} m\left(u^{2}+v^{2}\right)+m g y\)
Use the Chain Rule to compute E'(t) and show that E' (t) = 0 for all t \(\geq\) 20. Interpret the result.
ANSWER:Solution 53E
Step 1 of 3
Chain rule:
Let be a differentiable function of on its domain, where are differentiable functions of on an interval Then
