Aball is kicked into the air at time t=0. while the ball is in the air, its height above the ground after t seconds is -16t^2+40t feet. how many seconds after the ball is kicked will it hit the ground?

The amount of time it would take after the ball is kicked before it will hit the ground again is 2.5 seconds

The reason for the above time duration of the ball is presented as follows:

The known parameters of the ball kicked into the air are:

The time at which the ball is kicked into the air, t = 0

The equation that gives the height of the ball as a function of time, t, f(t), is presented as follows:

f(t) = -16·t² + 40·t

The required parameter:

The time it would take the ball to hit the ground

Method:

Equate the function for the height to zero, which is the height at ground level and solve for time, t as follows:

At ground level, we have:

h = f(t) = 0 = -16·t² + 40·t

Therefore, we get;

0 = -16·t² + 40·t

Factorizing gives;

0 = t × (-16·t + 40)

Therefore;

t = 0, or -16·t + 40 = 0

From which we have:

-16·t + 40 = 0

16·t = 40

t = 40/16 = 2.5

t = 2.5 seconds

The time it will take the ball to hit the ground after being kicked, t = 2.5 seconds

Learn more about time of flight of projectile motion here:

link

answer: all the values are within because it has a pattern that repeats from -6 to 4.

good luck!

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363