Answer to Question #166227 in Atomic and Nuclear Physics for nahom

Question

ABOY STANDING AN THE GRAND THRAWS A BALL STRAIGHT UP WITH AN INITIAL VELOCITY OF 12M/S.

A,WHAT IS THE MAXIMUM HEIGHT THE BALL REACHES?

B,HOW LONG DOES IT TAKE TO REACH THIS HEIGHT?

Answer 1

a), b) the dependence of height on time is

$h(t) = v_0t- \dfrac{gt^2}{2}.$

The maximum height corresponds to the zero velocity, so the time of flight is

$t_0 = \dfrac{v_0}{g}= \dfrac{12\,\mathrm{m/s}}{9.81\,\mathrm{m/s^2}} = 1.22\,\mathrm{s}.$

Therefore, the maximum height is

$h(t_0) = 12\,\mathrm{m/s}\cdot 1.22\,\mathrm{s} - \dfrac{9.81\,\mathrm{m/s^2}\cdot(1.22\,\mathrm{s})^2}{2} = 7.34\,\mathrm{m}.$