# A ball is tossed straight up from the edge of a 38 foot high building, with an initial upward velocity of 42 feet/second?

rotchm

a) You are asked to solve 0 = -16t^2+42t+38. You have seen the quadratic eqs right? Or how to factor? b) The vertex of the parabola -16t^2+42t+38 c) Again, The vertex of the parabola -16t^2+42t+38 Done! Btw, these questions are highschool questions. I say this because a few months back you (?) We doing limits, derivatives and IVT which are subjects beyond the current post. Have you forgotten how to work with parabolas?

llaffer: Part 1: solve for t when h(t) = 0

Part 1: solve for t when h(t) = 0: h(t) = -16t² + 42t + 38 0 = -16t² + 42t + 38 Divide everything by -2 to simplify and get rid of the negative in the high-degree coefficient: 0 = 8t² - 21t - 19 Quadratic equation: t = [ -b ± √(b² - 4ac)] / (2a) t = [ -(-21) ± √((-21)² - 4(8)(-19))] / (2 * 8) t = [ 21 ± √(441 + 608)] / 16 t = [ 21 ± √(1049)] / 16 We can't have a negative time, so throwing that out we get: t = (21 + √1049) / 16 or approx 3.34 (rounded to 2DP) sec To answer the other two, we can put this into vertex form: h(t) = a(t - h)² + k Where "h" will be the time of maximum height and "k" is that maximum height. h(t) = -16t² + 42t + 38 We want the right side to be in the form of (t² + bt) so we'll start with subtracting 38 from both sides then divide both sides by -16: h(t) - 38 = -16t² + 42t -[h(t) - 38] / 16 = t² - (21/8)t Completing the square by adding the square of (21/16) to both sides: -[h(t) - 38] / 16 + 441/256 = t² - (21/8)t + 441/256 Now we can factor the right side: -[h(t) - 38] / 16 + 441/256 = (t - 21/16)² Now solve for h(t) again. Subtract 441/256 from both sides: -[h(t) - 38] / 16 = (t - 21/16)² - 441/256 Multiply both sides by -16: h(t) - 38 = -16(t - 21/16)² + 441/16 And add 38 to both sides: h(t) = -16(t - 21/16)² + 441/16 + 38 h(t) = -16(t - 21/16)² + 441/16 + 308/16 h(t) = -16(t - 21/16)² + 1049/16 The maximum height is reached at 21/16 seconds (1.3125 seconds) with a height of 1049/16 ft (65.5625 ft)

david

-16t^2+42t+38 = 0 t^2- (21/8)t = 19/8 t^2- (21/8)t + 441/256 = 19/8 + 441/256 t - 21/16 = +/- sqrt [1049/256] t = 3.368 sec h' = -32t + 42 = 0 t = 42/32 = 21/16 <<< time to reach max height h(21/16) = 65.5625 <<< max height

Anonymous

< How long before it reaches its maximum height? > At the top of the arc its speed is zero. Change in speed = 42 ft/sec, time = 42 ft/sec / 32.2 ft/sec^2 = 1.30 sec < How high will it go? > vf^2 = vi^2 + 2 * a * d 0 = (42 ft/sec)^2 + 2 * -32.2 ft/sec^2 * d. Gravity operates opposite to the initial speed so it has the opposite sign. d = 27.4 ft. Since the building is 38 ft, the final height = 65.4 ft < How long before the ball hits the ground?(h(t))=0 > distance = 1/2 * a * time^2. At the top of the arc, its speed = 0 and the height is 65.4 ft 65.4 ft = 1/2 * 32.2 ft/sec^2 * time^2, time = 2.02 sec Time to reach the ground is the time to the top of the arc plus the time to fall to the ground, 3.32 sec