Question

7. From top of a tower an object is thrown up with 'u' m/s. When it falls back, it crosses the top of the tower with_____ m/s a) u/2g b) u²/g c) u/g d) u ​

Answer 1

Answer:

when body moves upward direction.

initial velocity =u

final velocity =v=0

now, v=u+at

Here, v=0,a=−g [acceleration is downward direction]

0=u−gt⇒t=u/g

Maximum height = h

u

2

=2gh⇒h=u

2

/2g---(1)

Case 2 :- when body moves downward direction.[Falling from maximum height]

initial velocity, u=0

final velocity = v

use formula, s=ut+1/2at

2

here,s=−h,u=0 and a=−g

−h=0−1/2gt

′2

−u

2

/2g=−gt

′2

/2 [from equation(1)]

t

′

=u/g

Hence, total time =t+t

′

=u/g+u/g=2u/g