Answer to Question #144447 in Quantum Mechanics for Kenneth

Question

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?

Answer 1

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

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