MATH SOLVE

5 months ago

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 [tex]h[/tex] with 0 in our equation, and solve for [tex]t[/tex]:

[tex]0=43-4t-5t^{2}[/tex]

[tex]-5t^{2}-4t+43=0[/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.

[tex]0=43-4t-5t^{2}[/tex]

[tex]-5t^{2}-4t+43=0[/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.