posted by xai on .
a soccer ball is kicked from the ground with an initial speed of 19.5 meter per second at upward of 45 degrees. a player 55 meter away in the direction of the kick starts running to meet the ball at the instant. what must be his average speed if he is catch the ball just before it hits the ground
It hits the ground
(Vo^2/g) = 38.8 meters away
The other player was
55-38.8 = 16.2 meters
away when it was kicked. He must run that distance before the ball hits the ground. That running time is
2*Vo*sin45/g = 2.814 seconds
Required average speed = 16.2/2.814 = 5.76 m/s
vertical velocity=19.5Cos45=13.79m/s. Time up=(v-u)/g=(0-13.79/-9.8=1.42s. T(up)+T(down)=2*1.42=2.85s. avrgS=D/T=55/2.85=19.30m/s.