Question

A body dropped top fall through 40 m during the last two seconds of its fall. The height of tower is (g I0 m/s2)​

Answer 1

Answer:-

h = 45 m

Solution:-

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

Answer 2

Step-by-step solution :

So in last 2 seconds, the height which ball covered was 40 m. Then it's speed was :

⇒ s = ut + 1/2 gt²

⇒ 40 = u(2) + 1/2 × 10 × 2²

⇒ 40 = 2u + 5•4

⇒ 40 - 20 = 2u

⇒ 10 m/s = u

This means that, Initial velocity here will become Final Velocity when thrown from top of house.

⇒ v² - u² = 2as

⇒ 10² - 0² = 2(10)s

⇒ 100/20 = 5 m

Total height = 40 m + 5 m = 45 m

Answer : 45 m