Question

A penny is dropped from the roof of a tall building. Find the following measurements; note initial velocity is 0 m/sec.

a) ___________velocity after one second

   ___________ velocity after two seconds

   ___________velocity after 3 seconds

b) ___________distance fallen after one second

   ___________distance fallen after two seconds

   ___________distance fallen after 3 seconds

2) A football is thrown downward at 18 m/sec.

a) What is the acceleration of the football as it falls?

b) What is the distance it falls after 0.3 seconds?

Answer 1

(a)

$$v=v_0-gt,$$ $$v(t=1\ s)=0-9.8\ \dfrac{m}{s^2}\cdot1\ s=-9.8\ \dfrac{m}{s},$$ $$v(t=2\ s)=0-9.8\ \dfrac{m}{s^2}\cdot2\ s=-19.6\ \dfrac{m}{s},$$ $$v(t=3\ s)=0-9.8\ \dfrac{m}{s^2}\cdot3\ s=-29.4\ \dfrac{m}{s}.$$

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

(b)

$$y=v_0t-\dfrac{1}{2}gt^2,$$ $$y(t=1\ s)=0-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(1\ s)^2=-4.9\ m,$$ $$y(t=2\ s)=0-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(2\ s)^2=-19.6\ m,$$ $$y(t=3\ s)=0-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(3\ s)^2=-44.1\ m.$$

The distances fallen after the 1, 2 and 3 seconds will be 4.9, 19.6 and 44.1 meters, respectively.

2) (a)

$$a_y=g=-9.8\ \dfrac{m}{s^2}.$$

(b)

$$y(t=0.3\ s)=-18\ \dfrac{m}{s}\cdot0.3\ s-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(0.3\ s)^2=-5.841\ m.$$

The distance it falls after 0.3 seconds equals 5.841 meters.