A body dropped from rest from the top of tower, falls half thetotal distance in last second of the fall. Find the time of flight and height of the tower
Answers
Answered by
3
OK
here is the best answer
In last 1sec it travels H/2 distance, at that instant let velocity be v
v^2=u^2+2g(H/2) | u=0
v=(gh)^1/2
Distance eq.
H/2= vt + (gt^2)/2 | t=1
H/2 = (gh)^1/2 +g/2
(H-g)^2 = 4gh
H^2 + g^2 =6gh
H=1.66 | 57.134 m
Time taken i.e. T
2H=g.t^2
T=(2H/g)^1/2
T=0.58 | 3.41 sec
hope u got it
here is the best answer
In last 1sec it travels H/2 distance, at that instant let velocity be v
v^2=u^2+2g(H/2) | u=0
v=(gh)^1/2
Distance eq.
H/2= vt + (gt^2)/2 | t=1
H/2 = (gh)^1/2 +g/2
(H-g)^2 = 4gh
H^2 + g^2 =6gh
H=1.66 | 57.134 m
Time taken i.e. T
2H=g.t^2
T=(2H/g)^1/2
T=0.58 | 3.41 sec
hope u got it
Similar questions