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Suppose a 10cm diameter 1kg ball is launched directly vertically with an unknown force but takes a known amount of time to return. Can we calculate the height it achieved?

I'm really not very mathematically-minded but I presume we can divide the time taken by 2 to get when the ball was at its zenith? So assuming it took 10 seconds from launch to landing then I presume it had 5 seconds of upward motion and 5 seconds of freefall? Are these presumptions correct? If so, it would seem that all I need to know is: from what height would a 1kg mass have been dropped if it takes 5 seconds to reach the ground?

Ignoring friction / terminal velocity I'm sure this must be a simple equation I just can't get my head around it. Your gentle guidance is appreciated ;)

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Yes, if we ignore air resistance then you can indeed divide the time by $2$, and the question is equivalent to "how far would an object dropped from rest fall in $5$ seconds?"

The acceleration in freefall (ignoring air resistance) is a constant $9.81$ metres per second per second, regardless of the mass or size of the object.

Let's use throughout the standard units of metres, seconds, metres per second and so on.

If you want to use a standard formula, we have this one for motion with constant acceleration:

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

Where $s$ is the displacement from the starting position, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time taken. Substituting $u = 0$, $a = 9.81$ and $t = 5$ gives:

$$s = 0 + (\frac{1}{2} \times 9.81 \times 5^2) = 122.625$$

Thus our answer is $123$ metres (to $3$ significant figures).

If you prefer a more intuitive approach, we could say that since the acceleration is constant, the average speed will be the speed after $2.5$ seconds, which is:

$$2.5 \times 9.81 = 24.525 \text{ (metres per second)}$$

And $\text{Distance} = \text{Average Speed} \times \text{Time} = 24.525 \times 5 = 122.625$

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  • $\begingroup$ Excellent answer. You seem to have quite accurately judged my mathematical ability and pitched your answer just right. Thank you! $\endgroup$ – Moob Feb 5 '18 at 21:58
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The simple answer to your question is to assume that we can ignore air resistance and the size and shape of the projectile. In which case, treating the situation as the motion of a particle moving freely under gravity, and assuming the acceleration towards the centre of the Earth is $9.8ms^{-2}$ then, using a standard equation of constant acceleration $$s=ut+\frac 12at^2\implies s=0+\frac 12(9.8)5^2=122.5m$$

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