The equation y = -4.9t^2 - 3.5t + 12.4 relates the height y (in meters) to the elapsed time t (in seconds) for a ball thrown downward at 3.5 meters per second from a height of 12.4 meters from the ground. In how many seconds will the ball hit the ground? Express your answer as a decimal rounded to the nearest hundredth.
+240 Just a quadratic, so we have,
\(-4.9t^2 - 3.5t + 12.4 = 0\)
Apply quadratic formula,
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(x = {3.5 \pm \sqrt{255.29} \over -9.8}\)
Simplify via calculator we get roots, \(x = -1.99, 1.28\). We want the positive root so the answer is 1.28
+240 Just a quadratic, so we have,
\(-4.9t^2 - 3.5t + 12.4 = 0\)
Apply quadratic formula,
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(x = {3.5 \pm \sqrt{255.29} \over -9.8}\)
Simplify via calculator we get roots, \(x = -1.99, 1.28\). We want the positive root so the answer is 1.28
