Answer to Question #151020 in Physics for Arvin Shopon

Question

The stone is thrown straight up from a height of 1.2 m at an initial speed of 11 m/ s. At what height and at what speed does the stone move 1.5 s after the moment of throwing?

Answer 1

a) We can find the height of the stone 1.5 s after the moment of throwing from the kinematic equation:

$$y = y_0 + v_0t-\dfrac{1}{2}gt^2,$$ $$y = 1.2\ m + 11\ \dfrac{m}{s}\cdot 1.5\ s-\dfrac{1}{2}\cdot 9.8\ \dfrac{m}{s^2}\cdot (1.5\ s)^2=6.67\ m.$$

b) We can find the speed of the stone 1.5 s after the moment of throwing from the kinematic equation:

$$v=v_0-gt=11\ \dfrac{m}{s}-9.8\ \dfrac{m}{s^2}\cdot 1.5\ s=-3.7\ m$$

The sign minus means that the speed of the stone directed downward.

Answer:

a) $y=6.67\ m.$

b) $v=3.7\ m,$ downward.