Question

The equation for the height (in meters) of a ball y = -4.9x2 +*+100, where x represents time (in seconds) after the ball is thrownApproximately when will the ball hit the ground?

Answer 1

The equation is given as

$y=-4.9x^2+x+100$

where y is the height and x is the time.

Note that at the point when the ball hits the ground, the height is 0.

Therefore, we equate y to 0, such that

$\begin{gathered} y=0 \\ \therefore \\ -4.9x^2+x+100=0 \end{gathered}$

Solving the quadratic equation using the quadratic formula,

$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

where

a = -4.9

b = 1

c = 100

Substituting into the equation, we have

$\begin{gathered} x=\frac{-1\pm\sqrt[]{1^2-(4\times\lbrack-4.9\rbrack\times100)}}{2\times1} \\ x=\frac{-1\pm\sqrt[]{1+1960}}{2} \\ x=\frac{-1\pm\sqrt[]{1961}}{2} \\ x=\frac{-1\pm44.28}{2} \end{gathered}$

Therefore, the value can be

$\begin{gathered} x=\frac{-1+44.28}{2} \\ x=21.64 \end{gathered}$

or

$\begin{gathered} x=\frac{-1-44.28}{2} \\ x=-22.64 \end{gathered}$

Since the time can only be positive, the time it takes for the ball to hit the ground is 21.64 seconds.