An arrow was shot vertically upward at a velocity of 50.0 m/s. What was its velocity after 2.00 seconds?
vf=vi-gt solve for vf
speed=distance/time
50.0/2.00= 25m/s
To find the velocity of the arrow after 2.00 seconds, we need to consider the effects of gravity.
The arrow is shot vertically upward, so we know that its initial velocity (u) is 50.0 m/s and its acceleration (a) due to gravity is -9.8 m/s^2 (since acceleration due to gravity is always directed downward).
We can use the following equation of motion to find the final velocity (v) after 2.00 seconds:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Plugging in the values we know:
v = 50.0 m/s + (-9.8 m/s^2) * 2.00 s
Calculating this expression:
v = 50.0 m/s + (-19.6 m/s)
v = 30.4 m/s
So, the velocity of the arrow after 2.00 seconds is 30.4 m/s.