Children playing in a playground on the flat roof of a city school lose their ball to the parking lot below. One of the teachers kicks the ball back up to the children as shown in the figure below. The playground is 5.30 m above the parking lot, and the school building's vertical wall is h = 6.50 m high, forming a 1.20 m high railing around the playground. The ball is launched at an angle of 0 = 53.0 above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (Due to the nature of this problem, do not use rounded intermediate values in your calculations-including answers submitted in WebAssign.) (a) Find the speed (in m/s) at which the ball was launched. 18.13 m/s (b) Find the vertical distance (in m) by which the ball clears the wall. 1.61 m (c) Find the horizontal distance (in m) from the wall to the point on the roof where the ball lands. X Can you find the time it takes the ball to travel to the final vertical height of the playground? Can you then relate this to the total horizontal distance the ball travels, and then the horizontal distance from the wall? m (d) What If? If the teacher always launches the ball with the speed found in part (a), what is the minimum angle (in degrees above the horizontal) at which he can launch the ball and still clear the playground railing? (Hint: You may need to use the trigonometric identity sec (0) = 1 + tan-(0).) X What is the final x-position? The final y-position? How much time does it take the ball to travel to the final x-position in terms of the angle? Using this result in the equation for the final y-position, you should be able to write a quadratic equation for tangent of the angle. Be sure to use the trigonometric identity in the hint. . above the horizontal (e) What would be the horizontal distance (in m) from the wall to the point on the roof where the ball lands in this case? X The approach you use should be identical to part (c), only now the initial angle is the value found in part (d). mGiven that : D = 24.0 m 0 = 53:0.0 15. 30 The play ground. $3,H- 5.30 m above the banking lot. tome, t= 2.205 Now, Ux = initial velocity along a direction = vase by = initial velocity Along 4 direction = using ax = acceleration along x direction = 0 ay 2 acceleration slang y direction = - q downwant ) It. takes 2.20 see to reach point vertically clove the wall: " RF + 2 2.2 Sec horizontal displacement of Ball, Sx= 24 m. ( a) for in direction.: 2 24 = Uxt + xox t past Bazalist onto 7 0 Ux = 824 2 . 2 2 ) U Cost = 24 2:2 24 2. 2 x Grs 530 2 18 . 13 . m / s : The speed At which the ball was launched, 830.U2 . 18.13. mis-:From ( a) we already get the speed at which the ball was launched, U= 18.13 m/s. Now,. when the ball lands on the roof, the vertical displacement of ball, sy = 5: 30 m . time taken at. Applying the urematic egan - sy = wy t + bayt? 2 5030 = ( 18 . 13 x Sm530 ) t - 5x9. 8xt2 = ) 4. 9t - 14.48 + + 5.3 = 0 4 t = 14.48 / (- 14. 48) 2- 4x 4. 9 * 5- 3 14.48 + 10. 29 2 x 4 . 8) 2x 4.9 14.48 + 10. 29 t / = and t2 2 14:48 - 10. 29 9 . 8 9. 8 2 2.53 see = 0.43 seefinal Here ball will be at a height of 5.3 m during downward journey al- to 2.53 see. The horizontal distance during 2:53 sec SX = Uxt 2 18:13 253 530 x 2.53 2 27. 6 m " The horizontal distance from the wall to the point on the roof where the ball lands is - ( 27 .6 - 24 ) m = 1316 Now , So = 24 m Sy = 6.50 m . Sy = Uyt + Lay t 2 - 6.5 = 18. 13 3mox t - 1x 9.8x t 2 - Co and Sx 2 Uxt SX 24 (ii) 18. 13 GO Pul- t in equa 6.5 = 18.13 x Smo x 24 - 2x9. 3 x 242 18.13 WSO ( 18.13 ) 2 CS20 6.5 = 24 tamo - 4. 9 x/ 24 78. 13 ) x Seco 24 tame - 8.6 sec20 4 6. 5 = 24+amD - 8.6 - 8.6 tome .: seco=1+tame] 2 8.6 tam D - 24 +amo + 15.1 = 0 * tano= 241 / ( 24 ) 2 4 x 8: 6 x 15.1 2x 8.6 . tamA - 241 7. 52 17- 2:. tanDi = 24+ 7.52 2 1. 83 2 01 = 61.35. 17 . 2 tam O2 2 24- 7.52 2 0.96 2 022 43.83 17 : 2 " The minimumn angle is at which he can launch the ball and still clean the playground falling is - 43.830 (e) when ball lands on roof vertical distance Sy = 5.3 m Sy = Uyt + Layt 2 5 . 3 2 18 . 13 x Sim ( 4 3 . 8 3 : ) t - 1 x 9. 8 x +2 7 5 .3 2 12. 56 + - 4.9+2 4. 9 " - 12.56 + + 5.3 20 t = 12 . 56 + (12 . 56 ) 2 4 x 4.9 * 5.3 2 x 4.9 12. 56 + 7.34 9. 8 tz 12. 56 + 7. 34 2 2.03 see 9. 8 to 2 12.56 - 7.34 20153 sec. 9 . 8 . Required time 2.03 See. . : Sox = Uxt 2 18. 13 x Cos ( 43 . 83 0 ) x 2 03 = 26. 55 m .: The horizontal distance from the wall would be = (26.55 - 24 ) M 2 / 2155 m