Question

Example Problem: Ball on a building A ball is thrown up with an initial velocity of 20.0 m/s at an initial height of 50.0 m above the ground. Determine: (a) The time at which the ball reaches its maximum height. (b) Thelmaximum height. (c) The time needed for the ball to return to its initial position. (d) The ball's speed when it returns to its initial position. (e) The time at which the ball hits the ground. (f) The velocity and speed of the ball at t = 5.00 s. 50.0 m

Answer 1

b) zero when a At maximum height relocity of the ball ball will be The standard requation of motion ball throwing upward v=o V = U-9 :, V24-29 H g=10m/s² U=20mis น uz H = ag 202 His manimum height - 2 x 10 20 m So The maximum height is som from throwing. point a) Again v=u.rgt ༡ s 10 6 The required to reach marimum height t.= 4 20 9 25 > The displacement s with initial velocity u is given by S=ut the gt² uzo, s=distance from manimum to initial point 2om ... 20 = 0 + xlo xť? 2 a: Z 4 =14 2 S ® Time of fall is 2S cs Scanned with CamScanner

20.2 S 62 - 50m าร 100 d) ?=42+ 29. u=o at top point v. is at initial point = 2x10 X 20 = 400 ... v = 20m/s at initial point e) the distance from its initial point to ground velocity ad initial print falling= 20m/s 5=ut + 1 gt² 50 = 204 + 1 x10xt? = 207 15th t2 = 10-4t :: +2+ 4t - 10-O in z -4 1 v 16-4(-10) 2 - 4 + 16440 3-48 ż (rre root not. posible) = 1:745 time required to hit ground = 1.74 S f)) orgt u = 20 m/s = 20 - 10x5 - 30 mis relocity is in in town dward direction so are sign. CS Scanned with CamScanner 2 va

point and Time required to reach manimum come back initial =45 distance Covered = 20+20= 40m Distance Covered in dow and in is s=ut + g t ² = 20x1+x10x1 U=20 m/s = 20 tis = 25 m Total distance 40+25 m time 5 S in speed 65 = 13 m/s 9)) From rectons low of addition O A + AB=OB ABB = 6 = = N A m 5 um ) 309 E W 19° From geometry d A S LA = 180°-30-Lo =180-30-90= 60° B Angle beth oh and AB = 120 120° d2=0A2+AB² +2 ou AB Cos 120 609 = 5?+42 +2: 5.4 (-5) = 25+16-20 21 id=458 m Angle of resultant o ß with oÀ $3120° X0-87 tano AB sind OA + A B CB¢ 5 + 4 (-0.5 1309 c = = 1-16 :.oa tuntee 40%o. 4X0-87 Oztan? (1.16) = 49.23 3 Resultom makes 49.23-30 19.23° with E. 6 please gire a rating. Comment doubt sure help. CS Scanned with CamScanner