"a ball is released from rest and falls a distance h under the influence of" gravity. what is its average speed during the fall?

square root of 1/2 gh verage being the sum of all the values of the velocity divided by the number of values divided by... Well... velocity is a continuous thing, so the sum becomes an integral and the number becomes a range. The velocity increases at a rate of g metres per second every second. This is a straight line in time. The integral of the velocity gives you the distance travelled, in this case h. So you have the integral part of the average. Now we need the time interval. How long does it take to travel the distance? Well, the velocity is the integral of the acceleration wrt time, which is just -gt + c (c is the initial velocity, in this case zero - the object falls from rest). The distance is the integral of velocity with respect to time, which is -(1/2)gt^2 + c (c is the initial height, in this case h). So the height of the object at any given time, y=-(1/2)gt^2 + h. We want the time, t_0, when the height is zero (once it has travelled a distance of h). 0 = -(1/2)gt_0^2 +h, rearranging we get that t_0 = sqrt(2h/g) So we know the integral of the velocity, it's h. And we know the time interval, it's just sqrt(2h/g). So we divide the integral by the interval to find that the average is h/sqrt(2h/g). Multiplying top and bottom by sqrt(2h/g) gives gh*sqrt(2h/g)/2h, which is just g*sqrt(2h/g)/2. Taking the g inside the sqrt gives the average velocity to be sqrt(2gh)/2