Question

A ball is thrown directly up in the air from a height of 5 feet with an initial velocity of 60 ft/s. Ignoring air resistance, how long until the ball hits the ground? Use the formula where h is the height of the ball in feet and t is the time in seconds since it is thrown. Round your answer to the nearest tenth.

Answer 1

Answer:

3.8 seconds

Step-by-step explanation:

Given equation

$h= -16t^2 + 60t + 5$

When the ball hits the ground then height is 0

So we replace h with 0 and solve for t

$0= -16t^2 + 60t + 5$

a= -16 , b= 60 and c= 5

Apply quadratic formula to solve for t

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

=$\frac{-60+\sqrt{60^2-4\left(-16\right)\cdot \:5}}{2\left(-16\right)}[/tex[tex]=\frac{-60+-\sqrt{3920}}{-32}$

$=\frac{-60+-28\sqrt{5}}{-32}$

$=\frac{4(-15+-7\sqrt{5})}{-32}$

$=\frac{(-15+-7\sqrt{5})}{-8}$

Now make two fractions and solve for x

t= $-\frac{7\sqrt{5}-15}{8}$=-0.0815

t= $\frac{7\sqrt{5}+15}{8}$=3.83

So answer is 3.8 seconds