Q:

# 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

A:
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.