What is the velocity just before the body touches the ground

[ubergewehr273]

Homework Statement

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 ?

Homework Equations

$v^2 = u^2 + 2as$

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.

 

[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'? :)

 

[ubergewehr273]

I get the answer as roughly 5 km/h.

 

[ehild]

I get the answer as roughly 5 km/h.

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

 

[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...

 

[ehild]

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...

It is wrong, The speeds do not add.

 

[NTW]

It is wrong, The speeds do not add.

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...

 

[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. v_f=v_i+gt, and h=v_it+(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.

 

[NTW]

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. v_f=v_i+gt, and h=v_it+(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.

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

 

[PeroK]

The second ball takes less time to drop from h (as it's traveling 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.