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).

Explanation:

Step1: Identify acceleration

In vertical - motion without air - drag, the acceleration due to gravity $a=- 9.8\ m/s^{2}\approx - 10\ m/s^{2}$. The negative sign indicates the direction is downward.

Step2: Use kinematic equations

The kinematic equations for vertical motion are $v = v_0+at$ and $\Delta y=v_0t+\frac{1}{2}at^{2}$, where $v_0 = 20\ m/s$ (initial vertical speed), $a=-10\ m/s^{2}$, $t$ is time.
When $t = 0$:
$v=v_0 + at=20+(-10)\times0 = 20\ m/s$
$\Delta y=v_0t+\frac{1}{2}at^{2}=20\times0+\frac{1}{2}\times(- 10)\times0^{2}=0\ m$
When $t = 1$:
$v=v_0+at=20+(-10)\times1=10\ m/s$
$\Delta y=v_0t+\frac{1}{2}at^{2}=20\times1+\frac{1}{2}\times(-10)\times1^{2}=20 - 5=15\ m$
When $t = 2$:
$v=v_0+at=20+(-10)\times2=0\ m/s$
$\Delta y=v_0t+\frac{1}{2}at^{2}=20\times2+\frac{1}{2}\times(-10)\times2^{2}=40 - 20 = 20\ m$

Answer:

t (s)	v (m/s)	$\Delta y$ (m)
1	10	15
2	0	20

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