If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: UP Class 11 Physics>Unit 3

Lesson 3: Kinematic equations for uniformly accelerated motion

# Free fall - 2 body solved numerical

Let's solve a numerical on 2 free falling bodies. One stone dropped and another one, thrown up, when and where do they meet? Created by Mahesh Shenoy.

## Want to join the conversation?

• how do we know a falling object on earth has a constant acceleration as stated by the narrator??
(1 vote)
• A falling object would generally mean a freely falling object on which no force apart from gravity acts to cause acceleration. And acceleration due to gravity is taken as a constant, g=9.8 ms^-2 (here it is taken as 10 for the sake of simplicity).

Well, to be accurate - acceleration due to gravity is actually not constant and varies with Height. But the variation is negligible and hence ignored in high school Physics calculations. You will learn more about this in Class 11.

:) HOPE THAT HELPED!
• What if we are asked to calculate the final velocity
(1 vote)
• Final velocity of the 1st stone is 10√20 m/s or 20√5 m/s (As u=0 m/s, a=10 m/s^2 and h=100 m, when we use the 2nd equation to find 't', we get it as √20 s or 2√5 s. Now if we use the 1st equation, we get v=10√20 m/s or 20√5 m/s).

Final velocity of 2nd stone is 0 m/s (because when an object is thrown against gravity its final velocity is 0 m/s).

Hope it helps!
(1 vote)
• I think that the answer in this video is correct, but the first equation is wrong. The acceleration due to gravity acts in the same direction for both the stones. He got it right because the other side of the equation was also wrong.