# Why is there a contradiction in this Newton's equations of motion related question? [closed]

One second after being thrown straight down, an object is falling with a speed of 20 m/s. How much distance does the object travel in the first second?

We took: $$v = 20\,{\rm m/s}$$ $$u = 0\,{\rm m/s}$$ $$t = 1\,{\rm s}$$ $$a = 10\,{\rm m/s}^2$$

I used Newton's second equation

$$s = ut + \frac{1}{2}at^2$$

and found the answer to be $5\,{\rm m}$.

Then I used Newton's third equation

$$v^2 - u^2 = 2as$$

and found the answer to be $20\,{\rm m}$.

EDIT 1(for Kyle and Chris): Consider someone is holding a ball exactly like this. Someone comes and hammers if with a lot of force from the top. Here, it has an initial velocity of 0 but attains higher speed much faster. So how can you say that it isn't possible to reach high velocity without applying jerk? I am "dropping from rest with a force".

• If the object was thrown downwards would $u=0$? – Chris2807 Apr 7 '15 at 22:34
• Of course. Why wouldn't it be? – Eisa Adil Apr 7 '15 at 22:46
• Do you know what $u$ is? Why do you think it's zero? – Javier Apr 7 '15 at 22:47
• If it was dropped, how did it get to be 20 m/s in one second? – Ryan Unger Apr 7 '15 at 22:50
• Bit of a contradiction in terms. You can't "drop from rest with a force". If you apply some force to the ball to give it an initial velocity you should apply the equations of constant acceleration at the end after the intial force has been applied. – Chris2807 Apr 7 '15 at 23:10

If the object was released from rest ($u=0\,{\rm m/s}$), what is its speed after $1\,{\rm s}$ if $a=10\,{\rm m/s}^2$?

Using:

$$v=u+at$$

you will find that the object was not released from rest...

You must solve for the objects initial velocity first:

$$v(t)=u+at\\ v(0)=u\\ v(1)=u+10\text{m/s}\\ =20\text{m/s}\\ u=10\text{m/s}$$

With this adjustment you should find the correct answer.

Well, the first detail you have to realise is that the object was thrown with some initial speed $v_0$, otherwise, the object would be travelling at approximately 10 m/s after one second. Because of this, $v_0$ should be 10 m/s. This was an ideal case though. In "dirtier" scenario, you would have to check by using:

$v_f = v_0 + at$ which in this case is,

$v_0 = (20) - (10)(1) = 10$ m/s

Using this,

$s = v_ot + \frac{1}{2}at^2$ so,

$s = (10)(1) + \frac{1}{2}(10)(1)^2 = 15$ m/s

and $s = \frac{v_f^2 - v_0^2}{2a}$ is $s = \frac{(20)^2 - (10)^2}{2(10)} = 15$ m/s

The detail was that you didn't realise that the ball was thrown with an initial speed of 10 m/s instead of just dropped. That should have been apparent with the 20 m/s after just 1 second information.

EDIT: Response to your edit. This is where the other definition of force, $F = \frac{dp}{dt}$ (ie a change in momentum), becomes useful. The smack is not a continuous force like gravity. It is a applied for a short time, subjecting the object to an acceleration for a short period. In the case of a smack, it is nearly instantaneous and the net result is a change in momentum that imparts an initial velocity.