\nExample 1<\/span>: Imagine a projectile launched according to the equation \nSolution<\/span>:\nh=-4.9(3.0)^2+22(3.0)+15=36.9m or 37m (sig digs)<\/p>\n \nYet, for an approximate height,\nh=-5.0(3.0)^2+22(3.0)+15=36m\n<\/p> \n36m is 1\/37, or 2.7% different from the exact answer. It gives a very helpful guide towards what to expect from the true answer. Moreover, using -5.0 instead of -4.9, the approximate answer is relatively easy to produce mentally.\n<\/p>\n \nKnowing very close to what the answer will be is helpful. Then, if a wrong keystroke is made calculating the exact answer, the student will be aware and try again.<\/p>\nJack of Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.\n","protected":false},"excerpt":{"rendered":" Tutoring physics, approximations often help. The tutor mentions a comparison between exact and approximate. Example 1: Imagine a projectile launched according to the equationh=-4.9t^2 +22t + 15 What would be its height at 3.0 seconds? Solution: h=-4.9(3.0)^2+22(3.0)+15=36.9m or 37m (sig …<\/p>\n
h=-4.9t^2 +22t + 15<\/span>\n\nWhat would be its height at 3.0 seconds?\n<\/p>\n