A ball is thrown from a height of 25 meters with an initial

A ball is thrown from a height of 25 meters with an initial downward velocity of 5m/s . The ball\'s height h (in meters) after t seconds is given by the following. h=255t-5t^2 How long after the ball is thrown does it hit the ground?

Solution

height is given by,
h=255t-5t^2

when it hits the ground, h= 0
put h=0,
255t-5t^2 = 0
5t^2 + 5t - 25 = 0
t^2 + t - 5 = 0
solve above quadratic equation,
t = {-1 +/- sqrt (1^2 - 4*1*(-5))} / (2*1)
   = {-1 +/- sqrt(21)} / 2
   = {-1 +/- 4.58} / 2
t can\'t be negative,
t = (-1 + 4.58)/2
   = 1.79 s

Answer: 1.79 s