Question

43. A body dropped from top of a tower fall
through 40 m during the last two seconds
of its fall. The height of tower is
(g = 10 m/s)
[CBSE AIPMT 1991)
(a) 60 m (b) 45 m (c) 80 m (d) 50 m

Answer 1

Answer:

Let h = the height of the tower that needs to be determined.

Let t be the time of fall.

Then (t - 2) would be the time to reach the top of the 40 meter mark, and let d be the distance fallen till it reaches the 40 m mark.

Using kinematics,

we can write: d = 1/2 g (t-2)²

And H = 1/2 g t²

H also = 40 + d

Then: H = 40 + 1/2 g (t-2)² = 1/2 g t²

Expand: 40 +5 (t² - 4 t + 4) = 5 t²

40 + 5 t² - 20 t + 20 = 5 t^2

- 20 t = - 60

t = 3 s

(t-2) = 1 second

In one second an object falls = 5 m

Then H = 45 m

☄Hope it helps☄

Answer 2

Answer:

The law of constant proportions is often referred to as Proust’s law or as the law of definite proportions. An illustration describing the mass ratio of elements in a few compounds is provided below. The ratio of the number of atoms of each element is provided below the mass ratio. For example, in a nitrogen dioxide (NO2) molecule, the ratio of the number of nitrogen and oxygen atoms is 1:2 but the mass ratio is 14:32 (or 7:16).