A stone is released from rest from height above the ground. it covers a distance 6m in last second of it's motion. find the height from which it was dropped.
Answers
Answered by
3
Sn =u+a/2(2n-1)
S6=10/2+2(2×n-1)
6=5+4n-2
6=3+4n
3=4n
n=3/4 second
So,the body takes 0.75 sec to reach the ground.
S=1/2gt^2
s=5×9/16
s=2.9 approximately
So, the body is thrown from the height of 2.9m.
This is the answer.
If you like mark me brainlest
S6=10/2+2(2×n-1)
6=5+4n-2
6=3+4n
3=4n
n=3/4 second
So,the body takes 0.75 sec to reach the ground.
S=1/2gt^2
s=5×9/16
s=2.9 approximately
So, the body is thrown from the height of 2.9m.
This is the answer.
If you like mark me brainlest
Answered by
3
u=0
H-6=g(u/g+1)/2
H-6=5
H=11m
6=0+5(2n-1)
6=10n-5
n=11/10 sec
Total height= 5(11/10)²
=121x5/100
=6.05m
Similar questions