Question

A child on a swing 1m above the ground at the lowest point and 6m above the ground at the highest point. The horizontal speed of the child at the lowest point of the swing is approximately.

Answer 1

Answer:

4.16m/s

Explanation:

under root 2gh formula

2×9.81×1 =4. approximately.

Answer 2

Answer:

v = 10m/s

Explanation:

loss in P.E = Gain in K.E

           mg(h1-h2)=1/2mv²

• let,

            m= 1

             lowest height=1m

             max height = 6m

Now,  

          (1)(10)(6-1) = 1/2(1)v²

                 50=1/2v²

                   50(2)=v²

                      100=v²

                        v²=100

                           v=√100

                              v=10m/s