Suppose that a ball is dropped from the upper observation deck of a building, 400 m above the ground. (a) What is the velocity of the ball after 6 seconds? (b) How fast is the ball traveling when it hits the ground? SOLUTION We will need to find the velocity both when t = 6 and when the ball hits the ground, so it's efficient to start by finding the velocity at a general time t = a. Using the equation of motion s = f(t) = 4.9t , we have the following. v(a) = lim f(a + h) - f(a) h - 0 h 4.9 - 4.9a2 = lim h - 0 h = lim 4.9(a + 2ah + h - a2) h 4.9 lim h - 0 h lim 4.9 h - 0 (a) The velocity, in m/s, after 6 seconds is v(6) = 9.8 Fm/s. (b) Since the observation deck is 400 m above the ground, the ball will hit the ground at the time t, in seconds, when s(t) = 400, that is, we have the following. 4.9t2 = 400 This gives the following. (Round your approximation to one decimal place.) 400 12 = - and 4.9