Question

a student drops a ball from a stationary helicopter. make a motion chart for the first 5s of the balls fall. neglect drag. round the acceleration to the nearest whole number (if necessary). use regular metric units (i.e. meters).

Explanation:

Step1: Determine acceleration

Near - Earth, the acceleration of a freely - falling object is $a = 9.8\ m/s^{2}\approx10\ m/s^{2}$.

Step2: Calculate velocity at each time

The formula for velocity of a freely - falling object is $v=v_0 + at$. Since $v_0 = 0\ m/s$, then $v = at$.
For $t = 1\ s$, $v=10\times1 = 10\ m/s$; for $t = 2\ s$, $v = 10\times2=20\ m/s$; for $t = 3\ s$, $v=10\times3 = 30\ m/s$; for $t = 4\ s$, $v=10\times4 = 40\ m/s$; for $t = 5\ s$, $v=10\times5 = 50\ m/s$.

Step3: Calculate displacement at each time

The formula for displacement of a freely - falling object is $\Delta y=v_0t+\frac{1}{2}at^{2}$. Since $v_0 = 0\ m/s$, then $\Delta y=\frac{1}{2}at^{2}$.
For $t = 1\ s$, $\Delta y=\frac{1}{2}\times10\times1^{2}=5\ m$; for $t = 2\ s$, $\Delta y=\frac{1}{2}\times10\times2^{2}=20\ m$; for $t = 3\ s$, $\Delta y=\frac{1}{2}\times10\times3^{2}=45\ m$; for $t = 4\ s$, $\Delta y=\frac{1}{2}\times10\times4^{2}=80\ m$; for $t = 5\ s$, $\Delta y=\frac{1}{2}\times10\times5^{2}=125\ m$.

Answer:

t (s)	v (m/s)	$\Delta y$ (m)
1	10	5
2	20	20
3	30	45
4	40	80
5	50	125