Question #123011

A projectile shot at an angle of 60° above the horizontal strikes a building 25m away at a point 16m above the point of projection. Find the magnitude and direction of the velocity of the projectile as it strikes the building

Expert's answer

Let the x axis be directed from the point of the start of the motion towards the base of the building and the y axis be perpendicular to the x axis and directed vertically.

The projection of the initial velocity onto x axis is "v_x = v_0\\cos\\alpha" and onto y axis is "v_y=v_0\\sin\\alpha - gt." Therefore, x coordinate will be "x=v_0\\cos\\alpha t, \\; y = v_0\\sin\\alpha t - \\dfrac{gt^2}{2}."

We may write the system

"\\begin{cases}\nx=v_0\\cos\\alpha t, \\\\\ny = v_0\\sin\\alpha t - \\dfrac{gt^2}{2}.\n\\end{cases}\n\\;\\;\\; \n\n\\begin{cases}\n25=v_0\\cos60^\\circ t, \\\\\n16 = v_0\\sin60^\\circ t - \\dfrac{9.8t^2}{2}.\n\\end{cases}\n\\;\\;\\;\n\\begin{cases}\nt=\\dfrac{25}{v_0\\cos60^\\circ}, \\\\\n16 = 25\\tan60^\\circ - \\dfrac{9.8\\cdot\\dfrac{25^2}{v_0^2\\cos^260^\\circ}}{2}.\n\\end{cases}"

Therefore, "v_0 = 21.2\\,\\mathrm{m\/s}, \\; t= 2.36\\,\\mathrm{s}." The velocity components will be

"v_x = v_0\\cos\\alpha = 10.6\\,\\mathrm{m\/s}, \\; v_y = v_0\\sin\\alpha - gt = -4.77\\,\\mathrm{m\/s}." So the total velocity will be "\\sqrt{v_x^2+v_y^2} = 11.6\\,\\mathrm{m\/s}" . The angle between the direction of x axis and the velocity is "\\tan\\beta = \\dfrac{v_y}{v_x} = -0.45, \\beta = -24^\\circ."

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