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 March 9th, 2018, 10:40 AM #1 Member     Joined: Jun 2017 From: Lima, Peru Posts: 89 Thanks: 1 Math Focus: Calculus Is there a better method for solving this problem of projectile motion? The problem states the following: A soccer player $\textrm{20.0 m}$ from the goal stands ready to score. In the way stands a goalkeeper, $\textrm{1.70 m}$ tall and $\textrm{5.0 m}$ out from the goal, whose crossbar is at $\textrm{2.44 m}$ high. The stricker kicks the ball toward the goal at 18 $\frac{m}{s}$. For what range of angles does the player score a goal meaning the ball passes above the goalkeeper but below the crossbar? To solve this situation I decided to use the equation for the trajectory which is deduced from motion in 2 directions. I'll skip that part to go straight to the question. (note: I defined as $\omega$ the launching angle and the rest is given as data from the problem $v_{0}=18\,\frac{m}{s}\,g=9.8\,\frac{m}{s^{2}}$ For this situation takes into account the distance of $\textrm{20 m}$ from the goal. $$y=\left ( v_{0}\sin\omega \right ) \frac{20}{v_{0}\cos\omega}-\frac{1}{2}g \left ( \frac{20}{v_{0}\cos\omega}\right )^{2}$$ Since it mentions the height of the crossbar is 2.44m then the above equation must be less than 2.44. $$\left ( v_{0}\sin\omega \right ) \frac{20}{v_{0}\cos\omega}-\frac{1}{2}g \left ( \frac{20}{v_{0}\cos\omega}\right )^{2}< 2.44$$ while for the other situation it would become into: Take into account the distance of $\textrm{15 m}$ from the goalkeeper: $$y=\left ( v_{0}\sin\omega \right ) \frac{15}{v_{0}\cos\omega}-\frac{1}{2}g \left ( \frac{15}{v_{0}\cos\omega}\right )^{2}$$ and by using the fact that the goalkeeper has a height of $\textrm{1.7 m}$ the above equation results into: $$\left ( v_{0}\sin\omega \right ) \frac{15}{v_{0}\cos\omega}-\frac{1}{2}g \left ( \frac{15}{v_{0}\cos\omega}\right )^{2} > 1.7$$ But as it can be seen there is a bit of a problem to solve both trigonometric equations manually or analitically? as how to find the value of the denominator moreover the tangent which is formed on the first term of both equations does not seem to be simplified with any algebraic manipulations. What would be the best way to get the angle? I already used Maple and I can solve the problem but my question arises on how to do it without plotting or using software or perhaps using a simple calculator. Any help please?.

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