I'm doing mechanics and I came across this question:
A ball of mass 0.2kg is dropped from a height of 2.5m above horizontal ground. After hitting the ground it rises to a height of 1.8m above the ground. Find the magnitude of the impulse received by the ball from the ground
The answer in the book I'm using says 2.59Ns. I first calculated the speed at which the ball hits the ground using v^2= u^2 +2as, which is 7m/s. The momentum when hitting the ground is therefore -1.4kgm/s.
The final speed is the bit which stumps me. If the answer really is 2.59Ns, then the speed has to be 5.9m/s. I successfully managed to compute this, when I solve the problem in terms of mgh and 1/2mv^2, however, I haven't actually covered that yet in the Mechanics course I'm doing, so I feel I shouldn't do it that way.
However, doing it using SUVAT, and assuming acceleration to be -9.8 and u as 7m/s, my final speed works out to be 3.7m/s, so clearly my impulse cannot be 2.59Ns. I then thought I should work out final speed using u as 0, which would make the total distance travelled 2.5-1.8=0.7m, and acceleration being 9.8, but I end up with the same answer. The only way I can see to get 5.9 as the speed, is to take u as 49 AND s as 0.7, although this isn't correct at all, as if we get the net distance, we can't then use any speed that wasn't the definitive initial speed.
So how do I calculate the final speed of this ball, to then calculate impulse, without having to go into GPE and Kinetic energy?