SOLUTION We will need to find the velocity both when t = 2 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) = 4.92, we have the following. v(a) = lim f(a + h) - f(a) h - 0 h = lim - 4.9a2 h - 0 = lim 4.9(a + 2ah + h2 - a2) h- 0 h 4.9 = lim h - 0 h = lim 4.9 (a) The velocity, in m/s, after 2 seconds is v(2) = 9.8 m / s . (b) Since the observation deck is 300 m above the ground, the ball will hit the ground at the time t, in seconds, when s(t) = 300, that is, we have the following. 4.9tz = 300 This gives the following. (Round your approximation to one decimal place.) 300 and 49 300 4.9 The velocity of the ball, in m/s, as it hits the ground is the following. (Round your answer to the nearest integer.) 300 300 v ( t ) = 4.9 = 9.8 4.9 m/s