A ball is thrown from a height of 43 meterswith an initial downward velocity of 4 m/s. The ball's height h (in meters) after t seconds is given by the following. h=43-4t-5t^2How long after the ball is thrown does it hit the ground?

Accepted Solution

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

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