# What's the highest height a ball thrown straight up will reach?

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This is more of a maths question than a physics question but it seems physics examiners prefer it and physics students hate it. So here we go:

Essential, assuming Newtonian mechanics (the ball is not going at the speed of light nor is it a tiny atom so Einstein can wait a bit), we are essentially dealing with a SUVAT question. So we need our SUVAT equations. There are five in total (learn them), but for our problem, we just need the one: v2 = u2 + 2as.

When the ball is at its highest height, just for a split, split second, it’s not moving. It’s not going up or down or anywhere so v = 0 (final velocity). Therefore our one equation becomes 0 = u2  + 2as.

As the question doesn’t say differently, let’s assume we’re on Earth. We don’t have to be but in this example, let’s say we are. Then we have a decision to make: which way is positive? This is because all of the SUVAT quantities are vectors so have a direction and something going up is definitely not going in the same direction as something coming down. Therefore, we need to make a choice, it doesn’t really matter to the maths but it’s useful for the physics. Usually, we take up as positive.

So, in that case, we’re on Earth and up is positive so our acceleration a = - 9.81. Acceleration is NEGATIVE! It acts downwards and we just decided that up was positive so it must be negative. Therefore we have u2  + 2as = u2  + 2(-9.8)s = u2  - 19.6s = 0 when we stick all of this into our equation.

So then we can do a bit of rearranging to get s = u2/19.6 for whatever our u (initial speed) is. The faster you throw it into the air, the higher it will go! And that’s that.

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