Answer to Question #166179 in Physics for Ymir gokson

Question

ball is thrown downward with an initial velocity of 20 ft/s. a) How fast is it moving 2s later? b) How far 

does 

 it fall in these time ?

Answer 1

(a)

$$v=v_0+gt=-20\ \dfrac{ft}{s}+(-32.17\ \dfrac{ft}{s^2})\cdot2\ s=-84.34\ \dfrac{ft}{s}.$$

The sign minus means that the velocity of the ball directed downward.

(b)

$$y=v_0t+\dfrac{1}{2}gt^2,$$ $$y=-20\ \dfrac{ft}{s}\cdot2\ s+\dfrac{1}{2}\cdot(-32.17\ \dfrac{ft}{s^2})\cdot(2\ s)^2=-104.34\ m.$$

The ball will fall 104.34 meters in these 2 seconds.