Answer to Question #107776 in Mechanics | Relativity for fatima

Question

Hamad rides his skateboard at a local skate park. He starts from rest at the top of the track as seen in the figure below and begins a descent down the track, always maintaining contact with the surface. The mass of the skateboard is negligible, as is friction except where noted.

(a) What is Shawn’s speed when he reaches the bottom of the initial dip, 18.0 m below the starting point?









(b) He then ascends the other side of the dip to the top of a hill, 10.0 m above the ground. What is his speed when he reaches this point? (2 mark)

Answer 1

a)

$$v=\sqrt{2gH}=\sqrt{2(9.8)(18)}=18.8\frac{m}{s}$$

b)

$$v=\sqrt{2g(H-h)}=\sqrt{2(9.8)(18-10)}=12.5\frac{m}{s}$$