# 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 ?

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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.)

`-4.9x^2+8x+1=0` Use the quadratic formula:

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:

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

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