Question

A rock is dropped off the edge of a cliff and its distance s (in feet) from the top of the cliff after t seconds is s(t) = 16t. Assume the distance from the top of the cliff to the water below is 400 ft. Answer part (a) and (b). (a) When will the rock strike the water? The rock will strike the water at 5 seconds. (Type an integer or a simplified fraction.) (b) Make a table of average velocities and approximate the velocity at which the rock strikes the water. Time Interval [4, 5] [4.5, 5] [4.9, 5] [4.99, 5] [4.999, 5] Average Velocity (ft /sec) (Type an integer or decimal rounded to three decimal places as needed.)

Answer 1

Given sit)=167² rock strike the water Сэт) (a) when ats distance from top of cliff is 400 ft i. s(t) = 400 =) 166 = 400 +2= = 400 252 t=5 16 - To find Rock will strikest 5 seconds (6) Average relocities formule; f(b)-f(a) over Caib b-a for (4,5]: f (5) -f( 4) 5-4 Ave range velocity = 1615) ² 1614) ² 400-256 - = 144 ..over(4153) |CS$canned with Camse Average velocity=1144

زهره ها) چه دی ها ) - (s)۴ دنيا - 5 2 (یا )276 (5) 6/ با32 - O0ها = 60 S کی ، م 15 2 (ا ) - (s) : 497) .ا - )6 (9) 2 )6(5) 2 - 00) با 1390 400 - 36016 (0) م - 8.4 ۱۱S و CSScanned with CamScanner

ever (+११,5] F(s) -4(4११) 5-4.११ - 16 (5)2-16(4.११) 0.01 = 400 - 398.4016 00 1159844 0.00 15 १.84 [+ १११, over F(s)- (4-१११) 5-4.१११ - 16 ()- 16(+ १११)2 0.001. 399.840016 59.१84 400 0.00 CS$canned with CamScanner 15१.१84

Time (415) (4.5,5) (4:9,5) (4.99,5) (4-999,5) anteral | 59.84 159.984 152 158.4 144 Average relocity cstanned with CamScanner