cosyn

R^2=u^2(2h/g) 5.76=u^2(4/10) u=sqrt(5.76/0.4)

since the p.e at the top of the hill was converted to k.e at the bottom:v^2=2gh 8^2=2*10h 64=20h h=3.2m

with newton 2nd law:when the jet release gas,impulse created result in an increase in the jet'svelocity:ft=mv=1.201*1.2103=1.454N/s 1.454=5.101(v-3.351) 0.285=v-3.351 v=3.636m/s

since the ball drop 100m above the ground,we calculate max height to get the vy at the point of landing:h=(100sin60)^2/20=375m so it travels(375m-100m) from the max height to the top of building vertically:v^2=2gh=2*10*275=5500 v=74.1m/s since the vx=100c...

on the y-component:fcos60+fcos60=1/2w 2fcos60=1/2w f=1/2w