Cannon Balls

We recieved solutions from Vladimir Radmanikov from South Island School, Samuel from Manor Park Community college and Laura from Millhaven school, who said the following:

If we take upwards as positive $x$ then this is motion under a constant downwards acceleration $a=-9.8$. We need to use the formula
$$
x(t) = v(0)t+\frac{1}{2}at^2= v(0)t-4.9t^2
$$
We want to know the starting speed $v(0)$ to make it hit the ground $x=0$ in 1s.
So, we must solve
$$
0 = v(0)\times 1 -4.9\times 1^2
$$
So, the ball must be shot up at $4.9ms^{-1}$


For other times we have
$$
v(0) = 4.9t
$$
So, double the time requires double the speed and so on. So, 10s travel time requires $49m^s{-1}$ etc., as in this table

Travel time	10s	100s	1000s	100,000s
Speed	$49ms^{-1}$	$490ms^{-1}$	$4.9kms^{-1}$	$490kms^{-1}$

Clearly the last one will go straight into space!

You used various method to find the higest point of the cannon ball, which occurred half way through the journey

Samuel said

We need to work out when the ball stops moving to find the highest point each time.Differentiating the first equation gives us
$$
v(t) = v(0) -9.8 t
$$
If $T$ is the time of maximum height then setting $v(T)=0$ gives $T = v(0)/9.8= t/2$. So it reaches the maximum height half way through the travel time in each case. Putting $t=5,50,500,500000$ into the first equation gives

Travel time	10s	100s	1000s	100,000s
Maximum height	122.5m	12.25km	1225 km	12250000 km

Laura noted

By conservation of energy the cannon ball must hit the ground with the same speed as it left the ground, so the up and down parts of the journey are mirror images. So the maximum height is reached exactly half way through the journey. So, at times 5s, 50s, 500s, 50000s.

Vladimir made use of another form of the kinematic equations

To find the maximum height I can use this formula
$$v(t)^2 = v(0)^2 + 2ax(t)$$
At the highest point $v(t)=0$, so
$$
x_{max} =\frac{v(0)^2}{2\times 9.8}
$$

This method also gives the values in the previous table

Vladimir though that, since the earths radius is 6000km the cannon ball taking 1000s was likely to return to earth, as this has a maximum height of 1225km, and was therefore still in range of gravity (even taking the decay of gravity into account). Laura noted that the escape velocity from earth is around 11 km per second, so all but the final cannon ball would return to earth.

Well done to you all