#16 - Derive the angle that a projectile must be launched in order to attain maximum horizontal distance. (Hint: find a relationship x($\theta$) then take dx/d$\theta$ to maximize distance)
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Step 1: The horizontal distance traveled by a projectile is given by the equation: $$x = v_0t\cos(\theta)$$ where $v_0$ is the initial velocity, $t$ is the time of flight, and $\theta$ is the launch angle. Show more…
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If a projectile is launched from the ground and air resistance is ignored, the object will travel farthest with an initial launch angle of 45 degrees or radians. However, if the object is launched from some point above ground level, the angle that yields a maximum horizontal distance of travel is not 45. The height of the path of such a projectile is modeled by the equation h = -16x^2 + ho, where h is the height, x is the horizontal distance the object travels, g is the acceleration due to gravity, and ho = 9 feet. We want to find the value for x that maximizes the horizontal distance traveled by the launched object. Let g = -32 feet per second per second, vo = 24 feet per second, and a is the initial height. Write an equation that describes the total horizontal distance traveled by the launched object at the moment when its flight is over and solve it for x. A) Write an equation that describes the total horizontal distance travel by the launched object the moment when its flight is over and solve it for x.
Kenny M.
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A small object is projected from level ground with an initial velocity of magnitude $16.0 \mathrm{~m} / \mathrm{s}$ and directed at an angle of $60.0^{\circ}$ above the horizontal. (a) What is the horizontal displacement of the object when it is at its maximum height? How does your result compare to the horizontal range $R$ of the object? (b) What is the vertical displacement of the object when its horizontal displacement is $80.0 \%$ of its horizontal range $R ?$ How does your result compare to the maximum height $h_{\max }$ reached by the object? (c) For when the object has horizontal displacement $x-x_{0}=\alpha R,$ where $\alpha$ is a positive constant, derive an expression (in terms of $\alpha$ ) for $\left(y-y_{0}\right) / h_{\max }$. Your result should not depend on the initial velocity or the angle of projection. Show that your expression gives the correct result when $\alpha=0.80,$ as is the case in part (b). Also show that your expression gives the correct result for $\alpha=0, \alpha=0.50$, and $\alpha=1.0$
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