Physics, asked by Darshitasarma122100, 1 year ago

A Stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s . Calculate when and where the two stone meet. ​

Answers

Answered by Anonymous
37

Solution:

Let the distance that the stone travelled from top to down be x.

(i) For stone dropped from the top of tower:

Given:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 10 m/s²

By second equation of motion,

\implies \boxed{\sf{s=ut+\frac{1}{2}gt^{2}}}

Substituting the values, we get:

\implies x = (0 × t) + ½ × 10 × t²

\implies x = 0 + 5t²

\implies x = 5t² - - - - (1)

(ii) For stone projected upwards:

Distance (s) = (100 - x)

Initial velocity (u) = 25 m/s

By second equation of motion,

\implies \boxed{\sf{s = ut + \frac{1}{2}gt^{2}}}

Substituting the values, we get:

\implies (100 - x) = (25 × t) + ½ × - 10 × t²

\implies 100 - x = 25t - 5t²

\implies - x = 25t - 5t² - 100

\implies - (x) = - (-25t² + 5t² + 100)

\implies x = - 25t² + 5t² + 100 - - - - (2)

From equation (1) and (2),

\implies x = x

\implies 5t² = -25t + 5t² + 100

\implies 5t² + 25t - 5t² = 100

\implies 25t = 100

\implies t = 100/25

\implies t = 4 s

Therefore:

Two stones will meet after 4 s.

Now, putting the value of t in eq (1),

\implies x = 5t²

\implies x = 5 (4)²

\implies x = 5 × 16

\implies x = 80 m

Hence:

The two stones will meet after 4 seconds when the falling stone has covered a height of 80 m.

____________________

Answered by: Niki Swar, Goa❤️

Similar questions