Question

make a motion chart for a cannonball launched with an initial speed of 20m/s. neglect drag.

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

a=-10 m/s²

t	v	δy
s	m/s	m
0	20	0
1	10	15
2	0	20
3	-10	15
4	-20	0
5	-30	-25
6	-40	-60

direction of motion at 3: up

Explanation:

Step1: Recall kinematic equations

The equations used are $v = v_0+at$ and $\Delta y=v_0t+\frac{1}{2}at^{2}$, where $v_0 = 20m/s$ and $a=- 10m/s^{2}$.

Step2: Calculate velocity at $t = 1s$

Using $v = v_0+at$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 1s$. So $v=20+( - 10)\times1=10m/s$.

Step3: Calculate displacement at $t = 1s$

Using $\Delta y=v_0t+\frac{1}{2}at^{2}$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 1s$. So $\Delta y=20\times1+\frac{1}{2}\times(-10)\times1^{2}=20 - 5=15m$.

Step4: Calculate velocity at $t = 2s$

Using $v = v_0+at$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 2s$. So $v=20+( - 10)\times2=0m/s$.

Step5: Calculate displacement at $t = 2s$

Using $\Delta y=v_0t+\frac{1}{2}at^{2}$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 2s$. So $\Delta y=20\times2+\frac{1}{2}\times(-10)\times2^{2}=40 - 20=20m$.

Step6: Calculate velocity at $t = 3s$

Using $v = v_0+at$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 3s$. So $v=20+( - 10)\times3=-10m/s$.

Step7: Calculate displacement at $t = 3s$

Using $\Delta y=v_0t+\frac{1}{2}at^{2}$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 3s$. So $\Delta y=20\times3+\frac{1}{2}\times(-10)\times3^{2}=60 - 45=15m$.

Step8: Calculate velocity at $t = 4s$

Using $v = v_0+at$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 4s$. So $v=20+( - 10)\times4=-20m/s$.

Step9: Calculate displacement at $t = 4s$

Using $\Delta y=v_0t+\frac{1}{2}at^{2}$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 4s$. So $\Delta y=20\times4+\frac{1}{2}\times(-10)\times4^{2}=80 - 80=0m$.

Step10: Calculate velocity at $t = 5s$

Using $v = v_0+at$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 5s$. So $v=20+( - 10)\times5=-30m/s$.

Step11: Calculate displacement at $t = 5s$

Using $\Delta y=v_0t+\frac{1}{2}at^{2}$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 5s$. So $\Delta y=20\times5+\frac{1}{2}\times(-10)\times5^{2}=100 - 125=-25m$.

Step12: Calculate velocity at $t = 6s$

Using $v = v_0+at$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 6s$. So $v=20+( - 10)\times6=-40m/s$.

Step13: Calculate displacement at $t = 6s$

Using $\Delta y=v_0t+\frac{1}{2}at^{2}$, substitute $v_0 = 20m/s$, $a=-10m/s^{2}$ and $t = 6s$. So $\Delta y=20\times6+\frac{1}{2}\times(-10)\times6^{2}=120 - 180=-60m$.

Answer:

The motion - chart is as follows:

$t(s)$	$v(m/s)$	$\Delta y(m)$
1	10	15
2	0	20
3	- 10	15
4	- 20	0
5	- 30	- 25
6	- 40	- 60