Question

a ball is launched upward from the ground at 30.0 m/s. its location at 1 - second intervals is shown. determine the time, velocity, and acceleration at the four indicated locations. enter negative values for any downward - directed vector.

time (s)	velocity (m/s)	accelen (m/s/s)
c	2		-9.81
d	3	0	-9.81
f	5	tap me!	-9.81

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Explanation:

Step1: Use velocity - time formula

The formula for velocity $v = v_0+at$, where $v_0 = 30.0$ m/s is the initial velocity, $a=- 9.81$ m/s² is the acceleration due to gravity.

Step2: Calculate velocity at $t = 2$s

Substitute $t = 2$s, $v_0 = 30.0$ m/s and $a=-9.81$ m/s² into the formula $v=v_0 + at$. So $v=30.0+( - 9.81)\times2=30.0 - 19.62 = 10.38$ m/s.

Step3: Calculate velocity at $t = 5$s

Substitute $t = 5$s, $v_0 = 30.0$ m/s and $a=-9.81$ m/s² into the formula $v=v_0+at$. So $v=30.0+( - 9.81)\times5=30.0-49.05=-19.05$ m/s.

Answer:

Time (s)	Velocity (m/s)	Acceleration (m/s²)
C: 2	10.38	- 9.81
D: 3	0	- 9.81
F: 5	- 19.05	- 9.81