Question

Sin 37° 0.60 Cos 37° 0.80 Tan 37° = 0.75 Plane of Fence Fence 20 m/s 2.5 m -32 m Note: Diagram not drawn to scale. A ball of mass 0.5 kilogram, initially at rest, is kicked directly toward a fence from a point 32 meters away, as shown above. The velocity of the ball as it leaves the kicker's foot is 20 meters per second at a angle of 37° above the horizontal. The top of the fence is 2.5 meters high. The kicker's foot is in contact with the ball for 0.05 second. The ball hits nothing while in flight and air resistance is negligible. (a) Determine the magnitude of the average net force exerted on the ball during the kick. (b) Determine the time it takes for the ball to reach the plane of the fence. (c) Will the ball hit the fence? If so, how far below the top of the fence will it hit? If not, how far above the top of the fence will it pass? (d) On the axes below, sketch the horizontal and vertical components of the velocity of the ball as functions of time until the ball reaches the plane of the fence.

Answer 1

@ Griven M = Mass of ball = 0. 5kg u = initial velocity - Oms - Velocity of ball after kick = 20 m/s O Velocity makes an angle o beth horizontal rg=h= height of fence - 2.5m. t = time of contact of foot coith ball 0.05s Applying impulse - momentum theoreom Froet & At = Mo & AP Eava * 0.05s = (0.5 KS) x 20 m 2017 2.5 Farg = - 10 kem/ 0.055 = 200 kB / la 32 » Favg - 200N cucine Teme taken beg ball to reach the plane of the fence - t - Horizontal distance = 52m = R ich Rot cose cox > ta 32m U coc 270 32m 20 COS 3 3 am 15,91272 .0034 S -> t = 2.0034 See © g yo+logt - t , po-o cas initialley ball ox on the goount) » g = 0 + 20 xsin (27°y x(@.00345) - 1x8 mke) (2.0034 5 - g = 24.11352 - 19.66669 ►9 = 4.44682 m 4.45m 4.45m 2.5m No, ball will not hit the fence. Aga gry= (4.45 -2.5) - 1.95m The ball tell Cross the fence 1.95 m above et W