Question

A ball is projected upward from the ground. Its distance in feet from the ground in t seconds is given by s left parenthesis t right parenthesis equals negative 16 t squared plus 126 t. At what times will the ball be 211 feet from the​ ground?

Answer 1

Answer:

t = 5.46 secs and t = 2.42 secs

Step-by-step explanation:

The distance of the ball from the ground is given by:

$s(t) = -16t^2 + 126t$

When the ball is at a distance 211 ft from the ground, that means s(t) = 211 ft. we need to find t:

$211 = -16t^2 + 126t$

$=> 16t^2 -126t + 211 = 0$

Using the quadratic formula:

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

where a = 16, b = 126 and c = 211

Hence:

$t = \frac{-(-126) + \sqrt{(-126)^2 - 4(16)(211)} }{2(16)}$     and          $t = \frac{-(-126) - \sqrt{(-126)^2 - 4(16)(211)} }{2(16)}$

$t = \frac{126 + \sqrt{15876 - 13504} }{32}$                     and           $t = \frac{126 - \sqrt{15876 - 13504} }{32}$

$t = \frac{126 + \sqrt{2372} }{32}$                                and           $t = \frac{126 - \sqrt{2372} }{32}$

$t = \frac{126 + 48.703 }{32}$                                and           $t = \frac{126 - 48.703 }{32}$

$t = \frac{174.703}{32}$                                     and           $t = \frac{77.297}{32}$

=> t = 5.46 secs and t = 2.42 secs

This means that the ball will be at 211 feet at two different times, first, after 2.42 seconds and then after 5.46 seconds.

This makes sense because the ball is projected upwards, which means that when going up, it attains 211 feet and also, when coming back down, it also attains 211 feet.