# What is the velocity just before the body touches the ground

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1. Nov 29, 2014

### Ashes Panigrahi

1. The problem statement, all variables and given/known data
A body is dropped from a height $h$ with an initial speed 0, reaches the ground with a velocity of 3 km/h. Another body of the same mass was dropped from the same height $h$ with an initial speed 4 km/h. The second body will reach the ground with what velocity ?

2. Relevant equations
$v^2 = u^2 + 2as$

3. The attempt at a solution
From the equation,
For the 1st body, $u=0$; $a=g=10m/s^2$,
Therefore $v=\sqrt{2gh}$
$\Rightarrow$ $h=0.0347 m$

For the 2nd body, $u'=4 km/h$; $a=g=10m/s^2$; $h=0.0347 m$,
Therefore $v'=\sqrt{u^2 + 2gh}$
$\Rightarrow$ $v'=\sqrt{1.23+0.7}$
where $v'$ is the final speed of 2nd body.

2. Nov 29, 2014

### ehild

It is correct. But you could have solved the problem easier.
The speed of the first body when reaching the ground is v1, and v12=2gh.
The speed of the second body at the ground is v', and v'2=u2+2gh= u2+v12.
You can keep the unit km/h. What do you get for v'? :)

3. Nov 29, 2014

### Ashes Panigrahi

I get the answer as roughly 5 km/h.

4. Nov 29, 2014

### ehild

Well, it is exactly 5 km/h. (With the easy method, v'2=32+42 :) )

5. Nov 29, 2014

### NTW

The first object, with zero initial velocity, reaches the ground at 3 km/h.

Now, as the second object has an initial velocity of 4 km/h, its final velocity will be those 4 km/h (that are preserved) plus the 3 km/h that gains in the fall => 4 km/h + 3 km/h = 7 km/h...

6. Nov 29, 2014

### ehild

It is wrong, The speeds do not add.

7. Nov 29, 2014

### NTW

They do... The velocities are added as vectors. In the problem, there are two velocities to add, the initial (I take it as vertical and downward, since the problem doesn't state any other thing) velocity of 4 km/h, and the final velocity resulting from the accelerated motion, that we know is, in this case, of 3 km/h. Hence, the two vectors are parallel and have the same direction, and in order to to get the resultant vector, it's enough to add their moduli...

8. Nov 29, 2014

### ehild

Starting from zero velocity, the first body reaches 3 km/h when falling down from height h.
The second body has 4 km/h initial velocity, so it needs a shorter time to do the same distance h than the first one. Its velocity does not increase by 3 km/h. vf=vi+gt, and h=vit+(g/2) t2 Eliminate t from the second equation and insert into the first one. You will see that the OP-s solution is correct.

9. Nov 29, 2014

### NTW

Yes, that's right... and I was wrong...

10. Nov 29, 2014

### PeroK

The second ball takes less time to drop from h (as it's travelling faster) so it accelerates for less time than the first ball. In general, if two balls fall for the same time, they will accelerate equally, but not if they fall the same distance.