ex1.5_17

$t$ $h=67-5t-4.9t^2$ . When does the ball hit the ground? Round your answer to the nearest hundredth.

The ball hits the ground when its height is zero. Therefore we must solve the equation

67-5t-4.9t^2=0.

Let's go straight to the quadratic formula. In this problem, we have

a=-4.9, \, b=-5, \, c=67.

It follows that

t = \frac{-(-5) \pm \sqrt{(-5)^2 -4(-4.9)(67)}}{2(-5)} = \frac{5 \pm \sqrt{1338.2}}{-10}.

The two solutions are

t = \frac{5 + \sqrt{1338.2}}{-10} \approx -4.16

t = \frac{5 - \sqrt{1338.2}}{-10} \approx 3.16.

Since the negative solution does not make sense in the context of the original problem, we throw it away and give only the realistic solution:

t \approx 3.16 \, \mbox{seconds}.