Answer to Question #239806 in Physics for Morgan Hopkin

Question #239806

A cannonball shot at 51 m/s at an angle of 38°. What will be the maximum height that the cannonball reaches? Units?


1
Expert's answer
2021-09-20T14:55:38-0400

The maximum height reached by the cannonball can be found as follows:


"y_{max}=\\dfrac{v_0^2sin^2\\theta}{2g},""y_{max}=\\dfrac{(51\\ \\dfrac{m}{s})^2\\cdot sin^238^{\\circ}}{2\\cdot9.8\\ \\dfrac{m}{s^2}}=50.3\\ m."

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