Question

launched vertically and with an avelocity of 100 from a p u nt the top of an o l ding. The on, infot, that the poter from the ground herecondse by (a) ) 0.55 and 5.7 and the city of the potato (in y 058 velocity 5.75 ) and (h) Find the speed of the potate in speed speed (e) Determine the time (ins) when the potato reaches samm height (d) of the potato sand 3.5 the acceleration in acceleration at 0.58 e) Determine how the potato is in the a round your answer to two decimal on Determine the velocity (in 1/4) of the potato upon hating the ground. (Round your answer to the

Answer 1

Solution partca)! = u-gt" 157.1581 Velocity at t=0.5's 236.457 96 = 32ft/s U = U= 100 (ft/s) - 32% 0.5 100 (4+15) - 16 10= 84 flls Velocity at t=5-755 - 122-9018 o= 1 = 10- 32x5.75 1o-184 v= - 84 ft/s

Part(b)! V(t) = ungt 0 : Ivo - 32x+ + Twis Potate will reach hence after time t= 3.125s it's maximum height. i Speed at t= 0.55 Sxt) = -167?+ $cot+do vodst? 0 - - 32+ +160+ 0 dt /art-o.5s = -32x0.5 +100 To= 8aftls speed at t= 5.75s SU) = -16 (3.125)2 + Im 43.125 +80 S.(t) = 236.25 ft But distance covered by potato 7 = 236-2547 - 45+ =) Y = 156-257 S, (t) = 1 gt2² - tz = (3.75 -3.125),

SCU = x 32 x 62.62572 S(t) = 110.25 ft Now total fine interval distance t= siass. the covered by Potato in s. S= 110.25 + 156.25 S=24:58 u= 266.5ff S.as(s) To 46.35 ft/s ] speed at (t: 0.55) = 1841 ft/s speed at 5.755 = 146.35] $+)s Part (c)! time when particle fotato reaches its max. height to Info t = 3.125s ] 4(d)! the particle acceleration a= d²st) dt of = -32St/s - Constanta arch at 0.55 = 32 545 395 - F32 fts

Pgot@) from the ground total height BA) = 9362557 Re ): 19 ? to 2 hou t *23525 36.25 - = 3.6425 3.04265 & T = 3.1255 total time = t+T = 3.8426 + 3.125 t = 6.96765 OR 116.975 Para ke : u= Sagh o= |2x 32 x 236:25 U = '5120 10 122.96 $ in ward