A 40 kg tiger leaps upwards from a flatform of 2.5m high with a speed of 8.70ms-1 at an angle of 40 degree

from the horizontal. what is its velocity when it hits the ground?

First look at what's happening in the y-direction. The tiger is jumping upwards with a velocity proportional to it's initial velcocity (Vo*sin40). It goes up, stops, and falls back down to the ground. Its total displacement is the height of the flatform (-2.5 meters, since it's going downward). Use the equation Vf^2=Vo^2+2ax (assuming gravitational acceleration of -9.8m/s^2).
Vf^2=Vo^2+2ax
Vf^2=(Vo*sin40)^2+2(-9.8)(-2.5)
Vf^2=(8.70*sin40)^2+2(-9.8)(-2.5)
Vf^2=91.022605243
Vf=9.540576777