Question

A youngster throws a rock from a bridge into the river 50 m below. The rock has a speed of 15 m/s when it leaves the youngster’s hand. Calculate the velocity of the rock when it strikes the water if it is thrown upward.

Answer 1

Assuming that the stone is thrown vertically... let's say it's a 1 kg stone.
It doesn't matter if it's thrown upwards or downwards as (assuming no air friction) it will pass the original throwing point with the same downwards velocity as it had upwards, 3 seconds previously. So it starts with 1/2 m v^2 = 0.5 * 1 * 15^2 = 112.5 J of ke
Then k.e. gained = gpe lost
k.e. gained = m g h = 1 * 10 * 50 = 500 J of Ke gained
so the final (total) ke is 612.5 J
which = 1/2 m v^2 = 0.5 v^2 here

so 0.5 v^2 = 612.5
so v^2 = 1225
so v = 35 m/s

Answer 2

We don't need to worry about the mass of the rock. It can be shown that a rock thrown vertically upward with an initial velocity, will have the same velocity when it returns to its starting point. Looking at the equation of motion \[v^2=u^2+2as\]we know that at the top of its flight(when s is maximum), v=0 and hence \(u^2=-2as\). On the way down, we know that u=0, and hence \(V^2=2as\) . But remember that acceleration is a vector, so its sign is opposite to the upwards case, and hence we can see that in this becomes \(V^2=-2as\) which is the same result above. What this means is that when you throw an object straight up with a particular velocity, it will return to the original point with the same velocity.

What this means for the problem at hand is that its vertical speed initial speed in the downward direction will be 15 m/s. Thus using again the equation \[v^2=u^2+2as\] and using a=10, and s =50, we find that the final speed will be \[v=\sqrt{u^2+2as}=\sqrt{15^2+2(10)(50)}=35\rm{m/s}\]