A ball is thrown into the air with an upward velocity of 80 feet per second. The function h = -16t^2 + 80t models the height h, in feet, of the ball at time t, in seconds. When will the ball reach the ground? after 3 seconds after 4 seconds after 5 seconds after 6 seconds

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Аноним Аноним · Answer 1 · 2018-11-25T16:03:51+01:00

The ball hits the ground when the height h(t) = 0 , so

-16t^2 + 80t = 0

-16t(t - 5) = 0

t- 5 = 0 or -16t^2 = 0 ( here t = 0 which corresponds to when ball is thrown)

t - 5 = 0 gives:-

t = 5 seconds (answer)

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erinna erinna · Answer 2 · 2019-10-09T09:28:49+02:00

Answer:

5 seconds

Step-by-step explanation:

The given height h, in feet, of the ball at time t, in seconds is defined by the function

[tex]h=-16t^2+80t[/tex]

The height of the ball is 0 if it reached on the ground.

Substitute h=0 in the given function.

[tex]0=-16t^2+80t[/tex]

[tex]0=-16t(t-5)[/tex]

Using 0 product property, we get

[tex]-16t=0\Rightarrow t=0[/tex]

[tex](t-5)\Rightarrow t=5[/tex]

It means ball will reach the ground at t=0(initial condition) and t=5.

Therefore, the ball will reach the ground after 5 seconds.