Find the average velocity of a particle released from rest from a height of 125 m over a time interval till it strikes the ground. Take g=10 m//s^2.
Answers
Answer:
s=ut + 1/2at^2
125=0+1/2×10×t^2
t=5seconds
as v=u+at
v at t=1 is 10m/s
v at t=2 is 20m/s
v at t=3 is 30m/s
v at t=4 is 40m/s
v at t=5 is 50m/s
thus avg v is 10+20+30+40+50/5
=150/5=30m/s
or u can find the average velocity by integrating directly from 0 to 5 sec
please mark as brainliest answer
Answer:
25 m/s
Explanation:
Sign convention : taking downwards direction as positive ( +)
s = 125m
g = 10 m/s^2
s = ut + 1/2at^2
=> 125 = 1/2 x 10 x t^2
=> 5t^2 = 125
=> t = +5 or t = -5
but for our purpose we're talking about the future and time cannot be negative.
=> t = 5 s
Average velocity = Total displacement / total time
= 125/5
= 25 m/s
Another way :
v^2 - u^2 = 2as
=> v^2 = 2 x 10 x 125
=> v^2 = 2500
=> v = 50 m/s
Since time and distance travelled are constant here,
we can use
avg. velocity = (u + v)/2
= (0 + 50)/2
= 25 m/s