Question

During a tennis match, a player serves the ball at 27.1 m/s, with the center of the ball leaving the racquet horizontally 2.31 m above the court surface. The net is 120 mway and 0.900 m high. When the ball reaches the met, (a) what is the distance between the center of the ball and the top of the net? (b) Suppose that, instead, the ball is served as before but now it leaves the racqueta 5.00 below the horontal. When the ball reaches the net, what now is the distance between the center of the ball and the top of the net? Enter a positive number if the ball clears the net. If the ball does not clear the net, your answer should be a negative number. Use g=9.81 (a) Number Units Units

Answer 1

→ SOL". Vore = 27.1 m/s Are 12 m 2.31n h 0.900 meter Soln (a) 9л this case where the ball i has velocity VO: (-so) 27.1 with m/5 angle below the Horizontal it to The time taken for the ball reach to net is given from t- Are An [2 Hore vocas (-s) 27.1 cas (-s) 27.1 cass' cass- 12 0.9961947 2701x 0.9961947 12 11.9543364 1.00 38 1984 El sec ICS Scanned with CamScanner

The vertical distance from the ground when the ball reaches the net e given by y= yot + voytt (19t") y = got Uosingot - ſgt² y 11 į x 9.8 m/s² (1) 2 27.1 Sin (-5') x I Sec 2.31m + sinc-o = - sino y 2.319 – 27.1 sin so x 1 sec - 5 x 9.6 x[1]? sinc-s] = -sias ac 2.317 - 27.1 x 0. 08715574 X 1 sec - 4.9 t a 4.9 y = 2.31 m - 2036 1 92063 0.0S1920 6 3 – 4.9 1 ar 4.95 1 9 2063 ot m y 11 - 4.951 The distance between the centre of the ball and top of net is ہز (-y) 0.900 (-4.954 ] CS Scanned with CamScanner w 158.5

time t = one 0.44280443 0.44 11 27.1 voie t=0.44 sec when the ball The vertical distance from the ground reaches the net is given by y = yo + voy t - 2 gt² - gt2 = 2.31 t rosinot [0] 2.31 + 0 £ 9,8x (1) 2 2:31 4.9 y a 2.5 g me fer The distance between the centre of the ball nef is and top af the 0.900 - (- 2:39) 3.49 meter CS Scanned with CamScanner