Question

A ball is thrown from a height of 43 meters

Answer 1

When the ball will hit the ground, the height will be zero. So we need to replace $h$ with 0 in our equation, and solve for $t$:
$0=43-4t-5t^{2}$
$-5t^{2}-4t+43=0$
To solve this equation we are going to use the quadratic formula: $t= \frac{-b(+/-) \sqrt{b^{2}-4ac} }{2a}$.
From our height equation, we can infer that $a=-5$, $b=-4$, and $c=43$. So lets replace those values in our quadratic formula to find $t$
$t= \frac{-(-4)(+/-) \sqrt{(-4)^{2}-4(-5)(43)} }{2(-5)}$
$t=- \frac{5}{8} - \frac{ \sqrt{713} }{8}$ or $t=- \frac{5}{8} + \frac{ \sqrt{713} }{8}$
$t=-3.96$ or $t=2.71$
Since time cannot be negative, $t=2.71$ is the solution of our equation.

We can conclude that the ball will hit the ground after 2.71 seconds.