Question

make a motion chart for a cannonball launched with an initial speed of 40m/s. neglect drag.

• round the acceleration to the nearest whole number (if necessary).
• neglect the initial height of the cannonball.
• after the ball lands on the ground enter zero for all speeds and heights (if necessary).
• use regular metric units (ie. meters).

a = m/s²

t	v	δy
0
1
2
3
4
5
6
7
8
9
10

direction of motion at 2s:

Explanation:

Step1: Determine the acceleration

The only acceleration acting on the cannon - ball in free - fall (neglecting air drag) is the acceleration due to gravity, $a=- 9.8\ m/s^{2}\approx - 10\ m/s^{2}$.

Step2: Use kinematic equations for velocity

The kinematic equation for velocity is $v = v_0+at$, where $v_0 = 40\ m/s$ and $a=-10\ m/s^{2}$.

Step3: Use kinematic equations for displacement

The kinematic equation for displacement is $\Delta y=v_0t+\frac{1}{2}at^{2}$, where $v_0 = 40\ m/s$ and $a=-10\ m/s^{2}$.

t (s)	v (m/s)	$\Delta y$ (m)
1	$40-10\times1 = 30$	$40\times1+\frac{1}{2}\times(- 10)\times1^{2}=40 - 5=35$
2	$40-10\times2 = 20$	$40\times2+\frac{1}{2}\times(-10)\times2^{2}=80 - 20 = 60$
3	$40-10\times3 = 10$	$40\times3+\frac{1}{2}\times(-10)\times3^{2}=120-45 = 75$
4	$40-10\times4 = 0$	$40\times4+\frac{1}{2}\times(-10)\times4^{2}=160 - 80 = 80$
5	$40-10\times5=-10$	$40\times5+\frac{1}{2}\times(-10)\times5^{2}=200 - 125 = 75$
6	$40-10\times6=-20$	$40\times6+\frac{1}{2}\times(-10)\times6^{2}=240-180 = 60$
7	$40-10\times7=-30$	$40\times7+\frac{1}{2}\times(-10)\times7^{2}=280 - 245 = 35$
8	$40-10\times8=-40$	$40\times8+\frac{1}{2}\times(-10)\times8^{2}=320 - 320 = 0$
9	0	0
10	0	0

The direction of motion at $t = 2s$: Since $v(2)=20\ m/s>0$, the direction of motion is upward.

Answer:

$a=-10\ m/s^{2}$; See the above - constructed table for values of $v$ and $\Delta y$ at different $t$ values; Direction of motion at 2s: Upward