The ball falls with a speed that is increasing at the rate of 10m/s every second.
If it takes t seconds to reach the ground, its speed will be 10t m/s.
Its average speed is (0+10t)/2=5t. Since h=vt where v is its average speed t=h/v=10/5t, 5t²=10 and t=√2 seconds. The speed on reaching the ground is 10√2 m/s.
In a perfectly elastic collision with ground the ball will leave the ground at speed 10√2, but will be slowed down by gravity.
If the ball and clay collide after T seconds the ball will have reduced its upward speed by gT m/s so it will be 10√2-10T=10(√2-T) and its momentum will be 40m(√2-T) upwards. The clay will be moving at speed gT downwards with momentum mgT=10mT. Total momentum before collision is -40m(√2-T)+10mT.
The clay will have fallen a distance 5T² and the ball will have reached a height of 10T√2-5T².
Note that the clay and the ball fall at the same rate, so when we add the two distances together we get 10T√2=h=10 and T=1/√2=√2/2 seconds corresponding to a distance 10-5/2=7.5 m from the ground.
Immediately before the collision, the speed of the clay is gT=5√2=7.07m/s. The speed of the ball is 10√2-5√2=5√2=7.07m/s.
The combined downward momentum before the collision is mgT-40m(√2-T)=5m√2-20m√2=-15m√2.
The clay-ball combo momentum after collision is (m+4m)V where V is the resultant velocity.
Because momentum is conserved, -15m√2=5mV so V=-3√2=4.24m/s. The resultant velocity is upward.