EXAMPLE 3 Suppose that a ball is dropped from the upper observation deck of a building, 350 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 tim t = a. Using the equation of motion s = f(t) = 4.9t, we have (a) = lim f(a +h) - f(a) h -0 h lim 4.9 2a - 4.9a2 X h lim 4.9(a2 + 2ah + h2 - a2) h lim 4.9 h-0 h lim 4.9 h-0 9.8a (a) The velocity after 6 s is v(6) = 9.8( 6 = 58.8 m/s. (b) Since the observation deck is 350 m above the ground, the ball will hit the ground at the time to when s(t; ) = 350, that is, 4.9t12 = 350. This gives the following. (Round your approximation to one decimal place.) t, 2 = 350 and 4.9 350 t1 4.9 X The velocity of the ball as it hits the ground is the following. (Round your answer to the nearest integer.) 350 v(t;) = 9.8t1 = 9.8 4.9 82.83 m/s