Let's first find the time that the projectile takes to reach its maximum height:
"v_y=v_0sin\\theta-gt,""0=v_0sin\\theta-gt""t=\\dfrac{v_0sin\\theta}{g}."Then, we can find the total flight time of the projectile:
"t_{flight}=2t=\\dfrac{2v_0sin\\theta}{g}."Finally, we can find the range of the projectile:
"x=v_0tcos\\theta,""x=v_0\\dfrac{2v_0sin\\theta}{g}cos\\theta=\\dfrac{v_0^2sin2\\theta}{g},""x=\\dfrac{(200\\ \\dfrac{m}{s})^2sin2\\cdot53^{\\circ}}{9.8\\ \\dfrac{m}{s^2}}=3923.5\\ m."
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