Asked by Anonymous
a stone is dtropped from the top of 40m heigh calculate its speed after 2 seconds .Also find the speed ith the stone strikes the ground.
Answers
Answered by
drwls
Let V be the velocity (positive downwards) and Y be the distance the stone has fallen (also positive downwards).
V = g t
g is the acceleration of gravity, which is 9.8 m/s^2
When it strikes the ground,
Y = 40 = (1/2) g t^2
t = sqrt (2 Y/g)= 2.857 s
V = sqrt (2 g Y) = g * t = 28 m/s
V = g t
g is the acceleration of gravity, which is 9.8 m/s^2
When it strikes the ground,
Y = 40 = (1/2) g t^2
t = sqrt (2 Y/g)= 2.857 s
V = sqrt (2 g Y) = g * t = 28 m/s
Answered by
Taiwo
200
Answered by
Akshara
Actually it is wrong....
S(distance)=40m
Speed= distance/time
i.e 40/2=20m/s.
S(distance)=40m
Speed= distance/time
i.e 40/2=20m/s.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.