IMG_1816

V(3)=—34(3)— L0 =—122feet/secc t=2.08/53 and —3.71254 v(t) =-32t-26 v(2.08753) =-32(2.08753)-26 =-92.801 feet/second 11.) To estimate the height of a building, a marble is dropped from the top of the building into a pool of calm water at ground level. a.) How high is the building if the splash is seen 5.6 seconds after the marble is dropped? s(t) =-161* +5, 0=-16(5.6) +s, 0=-501.76 +s, s, =501.76 The building is 501.76 feet high. b.) What is the average velocity of the marble on the time interval [1,4]? s(t) =—-16£* +501.76 s(4)-s(1) _ 245.76-485.76 4-1 3 = —80 feet/second high. 12.) A projectile is fired vertically upward with an initial velocity of 96 ft/sec from a tower 256 ft a.) How long will it take for the projectile to reach its maximum height? s(t) ==161* +96t +256 w(t) =-32t+96 -32t+96=0 32t =96 t =3seconds b.) What is its maximum height? $(3) =-16(3)* +96(3) + 256 =400 fect ¢.) How long will it take the projectile to reach its starting position on its way down? —161% +96t +256 = 256 -16* +96t =0 =161(t=6)=0 t=0,t=6 The projectile reaches its starting position after 6 seconds. d.) What is the velocity of the projectile when it passes its starting point on the way down? v(6) =-32(6)+96 =-96 feet/second ¢.) How long will it take for the projectile to hit the ground? s(t) =—-16" +96t +256 =0 —-16(* —6t-16)=0 —16(t-8)(t+2)=0 t=8t=-2 The projectile hits the ground at 8 seconds. f.) What will the speed of the projectile be when it hits the ground? |v(8)| =|-32(8) + 96[ =160 feet/second 13.) Uncle Si’s four-wheeler runs out of gas as it goes up a hill. The vehicle rolls to a stop then starts rolling backwards. As it rolls, its displacement d(¢) in feet from the bottom of the hill at # seconds since the vehicle ran out of gas is given by d(t)=145+31t ¢, a.) How far from the bottom of the hill was Uncle Si when he ran out of gas? When the four-wheeler runs out of gas, 7 = 0. d(0) =145+31(0)—(0)2 =145feet b.) When is his velocity positive? What does this mean in the context of the problem? v(t)=31-2¢ 0=31-2¢ t=15.5 The velocity is positive on the time interval (0,15.5).This means Si was moving up the hill from time 7 =0 to F=13.