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 (ie. meters).

a = m/s²

t (s)	v (m/s)	δy (m)
1
2
3
4
5

Explanation:

Step1: Identify acceleration

The acceleration due to gravity near the Earth's surface is approximately $g = 9.8\ m/s^{2}$. Rounding to the nearest whole - number, $a=10\ m/s^{2}$.

Step2: Use kinematic equations

The velocity - time equation is $v = v_0+at$, and the displacement - time equation is $\Delta y=v_0t+\frac{1}{2}at^{2}$, where $v_0 = 0\ m/s$ (starts from rest).

For $t = 0\ s$:

$v=v_0 + at=0+10\times0 = 0\ m/s$
$\Delta y=v_0t+\frac{1}{2}at^{2}=0\times0+\frac{1}{2}\times10\times0^{2}=0\ m$

For $t = 1\ s$:

$v=v_0 + at=0 + 10\times1=10\ m/s$
$\Delta y=v_0t+\frac{1}{2}at^{2}=0\times1+\frac{1}{2}\times10\times1^{2}=5\ m$

For $t = 2\ s$:

$v=v_0 + at=0+10\times2 = 20\ m/s$
$\Delta y=v_0t+\frac{1}{2}at^{2}=0\times2+\frac{1}{2}\times10\times2^{2}=20\ m$

For $t = 3\ s$:

$v=v_0 + at=0+10\times3=30\ m/s$
$\Delta y=v_0t+\frac{1}{2}at^{2}=0\times3+\frac{1}{2}\times10\times3^{2}=45\ m$

For $t = 4\ s$:

$v=v_0 + at=0+10\times4 = 40\ m/s$
$\Delta y=v_0t+\frac{1}{2}at^{2}=0\times4+\frac{1}{2}\times10\times4^{2}=80\ m$

For $t = 5\ s$:

$v=v_0 + at=0+10\times5=50\ m/s$
$\Delta y=v_0t+\frac{1}{2}at^{2}=0\times5+\frac{1}{2}\times10\times5^{2}=125\ m$

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

Answer:

$a = 10\ m/s^{2}$

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