Question

a ball is launched upward from the ground at 40.0 m/s. its location at 1 - second intervals is shown. determine the time, velocity, and acceleration at the five indicated locations. enter negative values for any downward - directed vector.

color - coding key:

check answers

the hardest working

time (s)	velocity (m/s)	accelen (m/s/s)
c	tap me!	tap me!	tap me!
d	tap me!	tap me!	tap me!
e	tap me!	tap me!	tap me!
g	tap me!	tap me!	tap me!

Explanation:

Step1: Identify the acceleration

The acceleration due to gravity near the Earth's surface is $a=- 9.8\ m/s^{2}$ (negative as it is downward - directed), and this is constant throughout the motion of the ball. So, for all locations, the acceleration is $-9.8\ m/s^{2}$.

Step2: Use the kinematic - equation for velocity

The kinematic equation for velocity is $v = v_0+at$, where $v_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.
For location C:
The initial velocity $v_0 = 40.0\ m/s$ and $a=-9.8\ m/s^{2}$. If we assume the time intervals are 1 - second intervals, and for location C, $t = 2\ s$.
$v=v_0 + at=40.0+( - 9.8)\times2=40.0 - 19.6 = 20.4\ m/s$
The time at C is $t = 2\ s$.
For location D:
$t = 3\ s$, $v=v_0+at=40.0+( - 9.8)\times3=40.0 - 29.4 = 10.6\ m/s$
The time at D is $t = 3\ s$.
For location E:
$t = 4\ s$, $v=v_0+at=40.0+( - 9.8)\times4=40.0 - 39.2 = 0.8\ m/s$
The time at E is $t = 4\ s$.
For location G:
$t = 5\ s$, $v=v_0+at=40.0+( - 9.8)\times5=40.0 - 49=-9\ m/s$
The time at G is $t = 5\ s$.

Answer:

Time (s)	Velocity (m/s)	Acceleration (m/s²)
C	2.0	20.4	- 9.8
D	3.0	10.6	- 9.8
E	4.0	0.8	- 9.8
G	5.0	- 9.0	- 9.8