Question

52. A body dropped from top of a tower falls through 40 m during the last two seconds of its
fall. The height of tower (g = 10ms) is given by

a) 60 m
b) 45 m
c) 80 m
d) 50 m​

Answer 1

Answer:

b) 45 m

Explanation:

let us suppose that during travelling 40m in last 2 sec

Its initial velocity was u.

Then, ut+1/2gt²=40 ( for last 2 sec)

⇒ 2u + 1/2 × 10 × 2^2 = 40

⇒ 2u + 20 = 40

⇒ u = (40 - 20)/2

⇒ u =  10 m/s

Now considering from top to this position,

⇒ u=0

Then by v²-u²=2gh

⇒ 10² - 0² = 2(10)s

⇒ 100/20 = 5 m

Total height = 40 m + 5 m = 45 m

The height of tower is 45 m