p08_024.pdf

2 gh where

v 0 y = v 0 sin 28 ◦ is the upward component of the skier’s “launch velocity.” To ﬁnd v 0 we use energy

conservation.

(a) The skier starts at rest y = 20 m above the point of “launch” so energy conservation leads to

mgy =

2 mv 2

⇒

v = 2 gy =20m / s

which becomes the initialspeed v 0 for the launch. Hence, the above equation relating h to v 0 yields

h =

( v 0 sin 28 ◦ ) 2

2 g

=4 . 4m .

(b) We see that all reference to mass cancels from the above computations, so a new value for the mass

will yield the same result as before.

24. From Chapter 4, we know the height h of the skier’s jump can be found from v y =0= v 0 y −