# A ball is thrown upward with an initial speed of 8 m/s. The equation h = -4.9x^2 + 8x + 1 models the height (h) in meters of the ball (x) seconds after it is thrown. Question: About how long will...

A ball is thrown upward with an initial speed of 8 m/s. The equation h = -4.9x^2 + 8x + 1 models the height (h) in meters of the ball (x) seconds after it is thrown. Question: About how long will it take for the ball to hit the ground ?

embizze | Certified Educator

Given the model `h=-4.9x^2+8x+1` , we want to know the x-value when h(x)=0. (The height will be zero when the ball hits the ground.)

If `ax^2+bx+c=0` then `x=(-b+-sqrt(b^2-4ac))/(2a)`

Here a=-4.9,b=8, and c=1:

`x=(-8+-sqrt(8^2-4(-4.9)(1)))/((2)(-9.8))`

So `x~~-.11666,x~~1.749` The negative value for x does not make sense in the context of the problem, so we use the positive solution.

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The ball hits the ground `(-8-sqrt(83.6))/(-9.8)` or approximately 1.7 seconds after the throw.

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The graph:

oldnick | Student

Indeed the quation provivded is an equation of motion of a body subject gravity.

Indeed if we derivate h(x) find the speed equation

`s(x)= -9.8x+8`  that x=0 gives  initial speed.

derivating again we have te costant  equation of acceleration

`a(x)=-9.8`

being motion equation :

`h(x)= 1/2 g x^2+v_0 x+h_0`

substituing  `g=-9.8`   `v_0=8`   `h_0=1`

we get:

`h(x)=-4.9x^2 +8x+1`

that si the equation provvided by the test.

Red line height, blue line speed, black line  acceleration.

Nothe tha speed decreases with positive value when  the ball  apporachs max height value ,wher becames 0. Afterword

assumes negative value( the ball comes down)

the speed is zero when:

`-9.8x+8=0`    `x=40/49`

the relaitve value is:  `4.265 ` mts ,that is max height value.

When the  ball hit the ground his speed is `9.1433` m/secs

oldnick | Student

The balls hit the ground when h=0

`-4.9x^2 +8x+1=0`

`x=(-8+-sqrt(64+19.6))/-9.8`

`x_1=-0.1166636286`

`x_2=1.7493166898`

The ball hit the ground after `1.7493166898`  secs fot the `x_1 ` isnt' accetable