an object dropped from a cliff falls with a constant acceleration of 10 m/s.find it's speed 5s after it was dropped.
jatinmittal1:
hey buddy its easy Apply 1st law of newton. v. = u+at. v = 0+10×5 =50m/s
very hard question sorry bro i don't know
Answers
Answered by
69
Hey!
______________
Initial velocity of the object (u) = 0
Acceleration (a) = 10 m/s^2
Time taken (t) = 5 s
Final velocity (v) = ?
We know,
v - u = at
v - 0 = 10 × 5
v - 0 = 50
v = 50 m/s
______________
Hope it helps...!!!
______________
Initial velocity of the object (u) = 0
Acceleration (a) = 10 m/s^2
Time taken (t) = 5 s
Final velocity (v) = ?
We know,
v - u = at
v - 0 = 10 × 5
v - 0 = 50
v = 50 m/s
______________
Hope it helps...!!!
awesome
:)
nice dude
Thanks
☺yo
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behna chaa gyi♡
why is this question featuring bin "brainleast questions" here? can anyone tell me ?
Well, i dont want it to be there. Actually, if a question is viewed 10+ times, that question gets selected as brainliest.
ok..thanks for clarification.
Answered by
49
hey mate
here is u r answer
u=0 (object dropped)
a= 10m/s
t= 5
v= u+at
= 0+10×5
50 m/s..
here is u r answer
u=0 (object dropped)
a= 10m/s
t= 5
v= u+at
= 0+10×5
50 m/s..
v=0.5at
how?
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