A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10m/s ,with what velocity will it strike the ground?after what time will it strike ground?

The equation that pertains to this is

V = AT (Equation 1)

A is the acceleration (10 m/sec), and T is the time (it has to be in seconds because the acceleration is in seconds).

The distance traveled is given by

D = V_0T + \frac 12 AT^2 (Equation 2)

where D is the distance, T is the time, A the acceleration, and Vo is the initial velocity. In your case, Vo = 0

20\,m = \frac 12 10\,\frac {m}{s} T^2

4\,s = T^2

T = 2\,sec. We now know that it takes 2 seconds. So, from equation 1 we get

V = 10\,\frac {m}{s^2} \,2\,s = 20\,\frac {m}{s}