Question

Problem 2: A ball is thrown from the top of a 50m building, with an initial velocity m Vi = 80 at an angle of 0 = 25º.

Answer 1

2) 80ins Problem 2: A ball is thrown from the top of a 50m building, with an initial velocity 1*25 vi = 80 - at an angle of 8 = 25º. a) [4 points) Find the time taken to reach the ground and the range (horizontal distance) to the point of the impact. 50m b) [7 points) Find the magnitude and direction of the velocity at the moment of impact. c) [2 points] If the ball does not bounce back, what is the energy dissipated from the impact? d) [5 points) If the ball does bounce back with the same angle as it hits the ground, what is the height of the 1st bounce, given the coefficient of restitution e = 0.8? e) [3 points] If the system only consists of the ball, what is the total work done on the system (from the moment the ball leaves the hand) at the highest point, and at the point of the impact? th Horizontal distance ance (R)= (Uxt => R=(8020525-> (8-24) =>/R=597.44 in a) considering vertical notion S,=-50 m U y= 80 singo SyzNytt I agte -50=(80 Sin 250) t- (9.8) 7 -> 4.92² -33.81 7 -50=0 2²-7t- 10.2=0 b) Vx: = 80 cosas Oz tant y Vy 80 sinas =10 - 32.92° t = 8.24 sec Vy= Uy tayt -19) (8.24) * Vy 46.94 mis y v = ſup?thy * v=86.4m's

o put mass (m) Energy disipated? K.E. when it reaches ground. =, (m) v² => I (1777(86-4)? y Energy disipated. (3 732.5) mi = 2 mis mass {mise ܢ ܢ 'g] d ) - 2 as e-0-8 => Ui=10.8) Vy = U,,= 37.55 mis výt vertical velocity after it after it bounces V'gina= uy = v' : 37:55 AV. 69.im's Sin 25 vi ²-uit =S'- (37.55 ) + is'- 71.95 mg 2(9.8) change in ball 늘 W2. Imre - x'm { (81.432 -18037] > TwoD= (532.5) J put mass (m) w.d.- M.E.