Question

A ball is thrown from an initial height of 1 meter with an initial upward velocity of 11 m/s. The ball's height h (in meters) after t seconds is given by the following.
h=1+11t-5^2
Find all values of t for which the ball's height is 6 meters.
Round your answer(s) to the nearest hundredth.

Answer 1

Answer:

$t=0.64$ and $t=1.56$ seconds

Step-by-step explanation:

$h=1+11t-5t^2\\\\6=1+11t-5t^2\\\\0=-5+11t-5t^2\\\\0=-5t^2+11t-5\\\\t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\t=\frac{-11\pm\sqrt{11^2-4(-5)(-5)}}{2(-5)}\\\\t=\frac{-11\pm\sqrt{121-100}}{-10}\\ \\t=\frac{-11\pm\sqrt{21}}{-10}\\\\t=\frac{11}{10}\pm\frac{\sqrt{21}}{10}\\ \\t_1\approx0.64\\\\t_2\approx1.56$

Therefore, the values of t for which the ball's height is 6 meters is $t=0.64$ and $t=1.56$ seconds.