I am studying motion in 1 dimension in kinematics and doing gravitation part. I was doing a question in which a ball is thrown upward from a tower of height $20\mathrm{m}$ with a speed of $5 \mathrm{m}/\mathrm{s}$, the question was to find the velocity just before the ball hits the ground. I am confused if I take whole equation i.e. displacement=$-20 \mathrm{m}$ and $g=-10\mathrm{m}/\mathrm{s}^2$ (as given in question) then what should I take as $u$? Is it $5+0$ or is it something else? Thanks for the help and have a great day.


1 Answer 1


The equation you mean to use is $v^2 = u^2 + 2a\Delta d$ where $u$ is the initial velocity.

Given your convention that upwards is positive, your initial velocity is $u=5\ \rm m/s$, since the ball is thrown upwards.

The displacement is downwards -- so according to your sign convention, it will be negative. Therefore, $\Delta d = -20 \ \rm m$.

Is Initial velocity of an object 0 after it reaches max height and starts free falling under 1 dimension?

When you throw the object up with a speed of 5 m/s, then the object will rise, attain a max height, and will then fall. Once it reaches the same height from which it was thrown, the object's speed will be 5 m/s again, except, this time, going downwards.

You can choose your "initial" point at any point you wish. That is, you can choose $u=+5 \ \rm m/s$, in which case the object goes up, stops, then falls back down after some time.

You could also choose $u=-5 \rm \ m/s$ as your initial velocity. Except, now, the object is directed downwards. This is fine. The key is to simply be consistent.

Next, you could even choose $u=0$. That's also fine. However, do note that your displacement $\Delta d$ is no longer -20 m. You chose your initial velocity to be zero, which means the starting position of your object is wherever the object has a velocity of zero, which is not 20 m above the ground (but higher).

In summary, if I tell you that an object:

  • at a position of $y=0\ \rm m$ has a velocity of $-10 \rm \ m/s$ and
  • at a position of $y=-15\ \rm m$ has a velocity of $-20 \rm \ m/s$

and I ask you for its velocity at a displacement of $-10 \rm \ m$, then both of these equations will give you the right answer: $v^2=\rm (-10 \ m/s)^2 + 2(-10 \ m/s^2)(-10 \ m)$ and $v^2=\rm (-20 \ m/s)^2 + 2(-10 \ m/s^2)(+5 \ m)$.

Hope this helps.

  • $\begingroup$ Thanks a lot for the help. $\endgroup$ May 28, 2021 at 16:49

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