Schaum's Outline of Theory and Problems of Beginning Calculus

CHAP. 18) RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY 139At a maximum value, qr turning point, U = 0. Hence,0 = 96 - 32t32t = 96t=3Thus, it takes 3 seconds for the rocket to reach its maximum height, which iss = 96(3) - 16(3)2 = 288 - 16(9) = 288 - 144 = 144 feet(c) When does the rocket of part (b) hit the ground?It suffices to set s = 0 in the free-fall equation (18J),0 = 96t - 16t20 = 6t - t2 [divide by 1630 = t(6 - t)from which t = 0 or t = 6. Hence, the rocket hits the ground again after 6 seconds.Notice that the rocket rose for 3 seconds to its maximum height, and then took 3 more seconds to fallback to the ground. In general, the upward flight from point P to point Q will take exactly the same time asthe downward flight from Q to P. In addition, the rocket will return to a given height with the same speed(magnitude of the velocity) that it had upon leaving that height.Solved Problems18.1 A stone is thrown straight down from the top of an 80-foot tower. If the initial speed is 64 feetper second, how long does it take to hit the ground, and with what speed does it hit the ground?Here so = 80 and uo = -64. (The speed is the magnitude of the velocity. The minus sign for uo indicatesthat the object is moving downward.) Hence,dss = 80 - 64t - 16t2 and U = - = -64 - 322dtThe stone hits the ground when s = 0,0 = 80 - 64t - 16t20 = t2 + 4t - 5 [divide by -1610 = (t + 5Xt - 1)t+5= 0 or t-1=0t=-5 or t=lSince the time of fall must be positive, t = 1 second. The velocity U when the stone hits the ground iso(1) = -64 - 32(1) = -64 - 32 = -96 feet per second15 5By (18.3), 96 feet per second = -(96) = 65 - miles per hour.22 1118.2 A rocket, shot straight up from the ground, reaches a height of 256 feet after 2 seconds. Whatwas its initial velocity, what will be its maximum height, and when does it reach its maximumheight?Since so = 0,dss= u,t - 16t2 and U =-= UO - 32tdt