Question

6. Mark has gotten stuck on the roof while cleaning the gutters. The dogs launch a Milky Way candy bar straight up but Mark fails to catch it. The candy bar passes Mark's outstretched hands at t;=0.69s after launch and again at t#1.685 (a) How high above the dogs' launch point is Mark? (b) What is the velocity of the candy bar at launch? (c) What is the velocity of the candy bar as it passes Mark on the way back down?

Answer 1

Time to = 0.69 see Time te) = 1.68 See Ahow motion equation My Mom P Hz) ha at ligte h = utilegt o dog ha utz- gth ® Now equofe epe Ø 4③ 7 uti - į gtr = utstegt → (u 90.6974329-44094 = fumer er ) - forgaten = 0.ggu =(4,19-21 * 1.52540-603) 7 O.ggu = 11:49687 u = 11.6404 ) 11.612 m/s) from eq? ha (11-613 x 069) 4 x981X60-69) 0. Th= 5.680m

> How high above the dogix. height in the 5.680 m 1 ③ vellity of candy bar! 74 = 11.613 mys ro o find rebreity of paxnes mark on the way bauten Vo = 4-9t1 = (1.013 - (9.81 *o-bo] =[41851 We = 4-ftz =(1-52 – 19ex 156] = - 4-851 48 41 Here die Ue are velveities of the candy bars As it paxsex mark on the way back down