KinematicsHard
Question
A body is released from the top of the tower H metre high. It takes t second to reach the ground. Where is the body after t/2 second of release?
Options
A.at 3H/4 metre from the ground
B.at H/2 metre from the ground
C.at H/6 metre from the ground
D.at H/4 metre from the ground
Solution
Applying S = ut = 
gt2 for the Ist case
H =gt2 .....(i)
Let H1 be the height after t/2 secs. So distance of fall + H = H1
H - H1 =
⇒ H - H1 =
gt2 .....(ii)
Dividing (i)and (ii),
⇒ 4H - 4H1 = H ⇒ H1 =
H
gt2 for the Ist caseH =
gt2 .....(i)Let H1 be the height after t/2 secs. So distance of fall + H = H1
H - H1 =
⇒ H - H1 =
gt2 .....(ii)Dividing (i)and (ii),
⇒ 4H - 4H1 = H ⇒ H1 =
HCreate a free account to view solution
View Solution FreeMore Kinematics Questions
A particle leaves the origin an initial velocity = 3 m/s and a constant acceleration = (-1.0î - 0.5ĵ) m/s2 Its...Apartile is thrown vertically upwards with a velocity of 19.6 ms-1. The position of the body after 4s will be...A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next 5 seconds and the...The velocity - time graph of the a body falling from rest under gravity and rebounding from the a solid surface is repre...A particle has an initial velocity and an acceleration of . Its speed after 10 s is...