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 cannonball.
• after the ball lands on the ground enter zero for all speeds and heights (if necessary).
• use regular metric units (i.e. meters).

Explanation:

Step1: Identify kinematic equations

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

Step2: Calculate velocity at $t = 1s$

$v=v_0 + at=20+( - 10)\times1=10m/s$.

Step3: Calculate displacement at $t = 1s$

$\Delta y=v_0t+\frac{1}{2}at^{2}=20\times1+\frac{1}{2}\times(-10)\times1^{2}=20 - 5=15m$.

Step4: Calculate velocity at $t = 2s$

$v=v_0+at=20+( - 10)\times2=0m/s$.

Step5: Calculate displacement at $t = 2s$

$\Delta y=v_0t+\frac{1}{2}at^{2}=20\times2+\frac{1}{2}\times(-10)\times2^{2}=40 - 20=20m$.

Step6: Calculate velocity at $t = 3s$

$v=v_0+at=20+( - 10)\times3=-10m/s$.

Step7: Calculate displacement at $t = 3s$

$\Delta y=v_0t+\frac{1}{2}at^{2}=20\times3+\frac{1}{2}\times(-10)\times3^{2}=60 - 45=15m$.

Step8: Calculate velocity at $t = 4s$

$v=v_0+at=20+( - 10)\times4=-20m/s$.

Step9: Calculate displacement at $t = 4s$

$\Delta y=v_0t+\frac{1}{2}at^{2}=20\times4+\frac{1}{2}\times(-10)\times4^{2}=80 - 80=0m$.

Answer:

The motion - chart values are as follows:

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