Answer to Question #144447 in Quantum Mechanics for Kenneth

Question #144447
A ball is thrown vertically downward , leaving the throwers hand with a speed of 10 m/s. What will be its speed after falling for 2 seconds? How far will it fall in 2 seconds?
1
Expert's answer
2020-11-16T07:48:52-0500

The velocity will be directed downwards during the motion and the acceleration is constant and equal to "g=9.81\\,\\mathrm{m\/s^2}" , so it can be calculated as

"v(t) = v_0 + g\\cdot t = 10\\,\\mathrm{m\/s}+ 9.81\\,\\mathrm{m\/s^2}\\cdot t"

After two seconds the velocity will be

"v(2\\,\\mathrm{s}) = 10\\,\\mathrm{m\/s}+ 9.81\\,\\mathrm{m\/s^2}\\cdot 2\\,\\mathrm{s} = 29.6\\,\\mathrm{m\/s}."


The distance can be calculated as

"s(t) = v_0(t) + \\dfrac{gt^2}{2} = 10\\,\\mathrm{m\/s}\\cdot2\\,\\mathrm{s} + \\dfrac{9.81\\cdot2^2\\,\\mathrm{s^2}}{2} = 39.6\\,\\mathrm{m}."


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