Home
Class 11
PHYSICS
A body falling from a high mimaret trave...

A body falling from a high mimaret travels 40 meters in the last 2 seconds of its fall to ground. Height of minaret in meters is (take `g=10(m)/(s^2)`)

A

60

B

45

C

80

D

50

Text Solution

Verified by Experts

The correct Answer is:
B
Height of minaret, `H = 1/2 g t^2`
In last 2sec body travels distance of 40m
`(H-40) = 1/2 g (t-2)^2`
`rArr H - 40 =1/2 g t^2 - 2gt + 2g`
`rArr H - 40 = H - 2g t + 2g`
`rArr t = 60/20 = 3sec`
`rArr H = 1/2 xx 10 xx (3)^2 = 45m.`
Promotional Banner

Similar Questions

Explore conceptually related problems

A body dropped from top of a tower falls through 40 m during the last two seconds of its fall. The height of tower in m is ( g= 10 m//s^@)

A body falling from a height of 10 m rebounds from the hard floor . It

A stone dropped from the top of a tower travels (5)/(9) th of the height of tower during the last second of fall. Height of the tower is: (Take g = 10 m//s^(2) )

If body travels half of its path in the last second of its fall from rest, find the time and height of its fall.

A stone falls from a cliff and travels 24.5 m in the last second before it reaches the ground at the foot of the cliff. Find the height of the cliff.

A ball rleased from the tope of a tower travels (11) /( 36) of the height of the tower in the last second of its journey. The height of the tower is (g= 10 ms^(2)).