timkle
Jun 19, 2009, 07:18 AM
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?
Perito
Jun 19, 2009, 07:28 AM
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}