Answer to Question #151423 in Physics for SHEN

Question #151423
Jane and Wussle are playing siatong on the field. They both agreed that the player with the farthest siatong to land will treat the loser with a snack. Jane hit her stick with an initial velocity of 12 m/s at 30° angle while Wussle launched his siatong with 15m/s at 25°

a. How far did wussle's siatong landed?
1
Expert's answer
2020-12-17T07:25:46-0500

a) Let's first find the time that Wussle's siatong takes to reach the maximum height:


"v_y=v_0sin\\theta-gt_{rise},""0=v_0sin\\theta-gt_{rise},""t_{rise}=\\dfrac{v_0sin\\theta}{g}."


Then, we can find the total time of flight of Wussle's siatong:


"t=2t_{rise}=\\dfrac{2\\cdot 15\\ \\dfrac{m}{s}\\cdot sin25^{\\circ}}{9.8\\ \\dfrac{m}{s^2}}=1.29\\ s."


Finally, we can find the range of Wussle's siatong from the kinematic equation:


"x=v_0tcos\\theta=15\\ \\dfrac{m}{s}\\cdot 1.29\\ s\\cdot cos25^{\\circ}=17.54\\ m."

Answer:

a) "x=17.54\\ m."


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