# Kinematic: Problem based on free fall

1. Mar 16, 2017

### NoahCygnus

1. The problem statement, all variables and given/known data
"A body dropped from the top of a tower covers a distance '7x' in the last second of its journey where 'x' is the distance covered in the first second. How much time does it take to reach the ground ?"
2. Relevant equations
$S = ut - \frac{1}{2}gt^2$

3. The attempt at a solution
Displacement in the first second;
$-x = 0 - \frac{1}{2}gt^2$
$x = 4.9 m$
Displacement during last second;
$7x \longrightarrow 7(4.9) \approx 35 m$

That's all I could do. I don't know the distance covered between first second and last second , so I don't know how to utilise the current information I've to find the time .

2. Mar 16, 2017

### cnh1995

What is the velocity of the body at the start of this interval? What is the velocity of the body at the end of this interval?

3. Mar 16, 2017

### NoahCygnus

Initially , it is 0 as the body is dropped from rest.
In the last second , I've no idea what the initial velocity is. Considering that the body hits the ground in the last second , it should come to a stop , doesn't that mean the final velocity should be 0? I'm not sure about it. If that's the case should i be able to find the initial velocity of the last second using the equation $v^2 - u^2 = 2as$ ?

4. Mar 16, 2017

### cnh1995

No.
In other words, what is the velocity of the body at t=1s? What is the velocity of the body 1 second "before" it hits the ground?

You need to use relevant kinematic equations involving time.

5. Mar 16, 2017

### NoahCygnus

At t= 1,
$v = u - gt$ as u = 0 , t = 1 , it should be ;
$v= 9.8 m/s$

1 second before it hits the ground , I don't know how to get that. Help me out Sheldon.

6. Mar 16, 2017

### cnh1995

You know the displacement and time in this interval. Which equation contains these two terms with unknown initial velocity?

7. Mar 16, 2017

### NoahCygnus

So the time interval is 1 second, using the equation
$-35 = u(1) - 5(1)$
$u = -30 m/s$

8. Mar 16, 2017

### cnh1995

Yes. 30m/s downwards.

Now you have initial and final velocities in that interval. You can calculate the time from that.

9. Mar 16, 2017

### NoahCygnus

Final velocity should be
$v = -30 - 9.8$
$v \approx -40$
Now to find the total time.
$-40 = 0 -9.8t$
$t = 4 secs$
Right ?

10. Mar 16, 2017

### cnh1995

Right.
(Be consistent with the value of g. Use either 9.8m/s2 or 10m/s2).

11. Mar 16, 2017

### NoahCygnus

Thank you . And you said the final velocity won't be 0. I don't get it , does the body hit the ground at the last second. if so shouldn't the body come to a stop and final velocity be 0? Or you don't take into account the contact force ground will exert and don't even consider its there ?

12. Mar 16, 2017

### cnh1995

The body hits the ground with some non-zero velocity.
Its velocity may become zero "after" it hits the ground (or it might bounce off) and that gets you into elastic-inelastic collisions. But whatever happens "after" the collision is a different issue. You only need to consider the time between the instant at which the ball is dropped and the instant at which it hits the ground. Only this interval describes the free fall of the body.