Question

A ball is thrown into the air with an upward velocity of 80 feet per second. The function

Answer 1

Answer:

After 5 seconds

Step-by-step explanation:

One is given the following expression to model to the flight path of a ball:

$h=-16t^2+80t$

In this problem, one is asked to find the time at which the ball lands on the ground. The easiest way to do so is to factor the expression. Take out factors that both terms have in common. The one will set the equation equal to zero, as this is the height at which one wants to find the time for. Finally, use the zero product property to solve. The zero product property states that any number times zero equals zero. Apply these ideas to the given function:

$h=-16t^2+80t$

Factor the expression; both terms have a common factor of (-16t):

$h=-16t^2+80t$

$h=-16t(t-5)$

Set the equation equal to zero to solve for when the ball has a height of zero or rather lands on the ground:

$h=-16t(t-5)$

$0=-16t(t-5)$

Now use the zero product property, this states that any number times zero equals zero. Set each of the factors equal to zero, and solve to find the value of (t) for which the factor will equal zero:

$0=-16t(t-5)$

$0=-16t$                          $t - 5 =0$

Solve for the value (t) using inverse operations,

$0=-16t$                            $t - 5 =0$

(divide by (-16)                   (add (5))

$0=t$                                     $t=5$

As one can see, the ball started on the ground, therefore at (0) seconds, its height was (0). Then, after (5) seconds the ball landed on the ground, therefore, after (5) seconds its height was again (0).