Step 1 :We are given a ball thrown from a height of 40 meters with an initial downward velocity of 5 m/s. The height of the ball after t seconds is given by the equation \(h=40-5t-5t^{2}\). We need to find the time when the ball hits the ground, i.e., when \(h=0\).
Step 2 :Setting \(h=0\) in the equation, we get the quadratic equation \(0=40-5t-5t^{2}\). This can be rewritten as \(5t^{2}+5t-40=0\).
Step 3 :We can solve this quadratic equation using the quadratic formula \(t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\), where \(a=-5\), \(b=-5\), and \(c=40\).
Step 4 :Substituting the values of \(a\), \(b\), and \(c\) into the quadratic formula, we get \(t=\frac{5 \pm \sqrt{(-5)^{2}-4*(-5)*40}}{2*(-5)}\).
Step 5 :Solving this, we get two solutions for \(t\), \(t1 = -3.37\) seconds and \(t2 = 2.37\) seconds.
Step 6 :However, time cannot be negative, so we discard the negative solution. Therefore, the ball hits the ground after approximately 2.37 seconds.
Step 7 :Final Answer: The ball hits the ground after approximately \(\boxed{2.37}\) seconds.