Aball is released at the top of a ramp at t =0. which is the speed of the ball at t=4

Accel;eration due to gravity= 9.8m/s^2. Initial velocity = 0(given). Time=4. Applying kinematic equation v=u+at, v= 0 + 9.8x4 = 39.2. Therefore speed of the ball after t=4 is 39.2

AbhiGhost your answer would be right if we were talking about free fall. The force that is providing the acceleration is Fx=mgsinФ, where Ф is the angle the ramp makes with the horizontal.
Fx=ma=mgsinФ
Hence:
a=gsinФ.
And the speed is v=a*t=gsinФ*t.