A ball is thrown from a height of 49 meters with an initial

A ball is thrown from a height of 49 meters with an initial downward velocity of 2 m/s. The ball\'s height h (in meters) after t seconds is given by the following. h = 49 - 2t - 5t^2 How long after the ball is thrown does it hit the ground? Round your answer(s) to the nearest hundredth. (If there is more than one answer, use the \"or\" button.)

Solution

We have h = -5t² -2t + 49 = -5( t² + 2t/5 - 49/5) = -5 [ t² + 2t/5 + 1/25 - ( 49/5 + 1/25)] = -5( t+1/5)² + 5( 245+1)/25 i.e. h = -5( t+1/5)² + 246/5. The is the equation of a parabola, in vertex form, which opens downwards. When the ball hits the grpond, h = -49,so that - 49 = -5t² -2t + 49 or, 5t² + 2t - 98 = 0. On using the quadratic formula , we have t = [ -2 ± { 2² - 4*5*(-98)} ] / 2*5 = [ -2± (4+1960)]/10 = [-2± 1964)] /10 = (-2 ± 44.41703961)/10 . Since t cannot be negative, we have t = (-2 + 44.41703961)/10 = 42.41703961/10 = 2.241703961 = 2.24 seconds ( on rounding off to the nearest hundredth).

Note: The solution to ax² +bx + c = 0 is x = [-b ± (b² -4ac)]/2a . This is the quadratic formula.