Answer to Question #106025 in Mechanics | Relativity for Ella

Question #106025
A ball is kicked such that it leaves the ground at an upwards angle with an initial velocity of vi.
The ball takes 3.1 s to hit the ground 26 m from where it was kicked.
What is the magnitude of the initial velocity of the ball? (in m s−1 to 2.s.f)
(note: ignore the effects of air resistance when solving this problem)
Expert's answer


"t_0=3.1 \\mathrm{\\ s} \\\\\nL=26 \\mathrm{\\ m}"

Find: "v_0".


Coordinate on the y-axis can be expressed as:

"y(t)=\\int\\int a_y dt dt = \\int\\int -g dt dt = \\int(v_{0y} - gt)dt = y_0+ v_{oy}t - \\cfrac{gt^2}{2}"

It is known that the initial y-coordinate "y(0)=y_0=0" m and it will be the same at the time "t_0", so:

"v_{0y}t-\\cfrac{gt_0^2}{2}=0\\Rightarrow v_{0y}=\\cfrac{gt_0}{2}"

Coordinate on the x-axis can be expressed as:

"x(t)=\\int \\int a_xdtdt=\\int v_{0x} dt=x_0+v_{0x}t"

It is known that the initial x-coordinate "x(0)=x_0=0" and it will be equal to "L" at the time "t_0", so:

"v_{0x}t_0=L\\Rightarrow v_{0x}=\\cfrac{L}{t_0}"

Thus, according to the Pythagorean theorem the initial velocity "v_0" is equal to:

"v_0=\\sqrt{v_{0x}^2+v_{0y}^2}=\\sqrt{(\\cfrac{L}{t_0})^2+(\\cfrac{gt_0}{2})^2}=\\sqrt{\\cfrac{L^2}{t_0^2}+\\cfrac{g^2t_0^2}{4}}=\\sqrt{\\cfrac{26^2}{3.1^2}+\\cfrac{9.8^23.1^2}{4}}\\approx 17.35\\mathrm{\\ m\/s}"

Answer: "v_0\\approx17.35 \\mathrm{\\ m\/s}."

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