If u could help me!! I'm a very slow student

Full question:

Yolanda throws a sphere up in the air from the top of a building. After tt seconds, the height of the sphere above the ground is hh feet, where** h=−16t2+30t+78.h=−16t2+30t+78.** After how many seconds does the sphere hit the ground?

beebee1 Jan 3, 2018

#2**+1 **

After t seconds, the height of the sphere above the ground is h feet, where

h = -16t^{2} + 30t + 78

After how many seconds does the sphere hit the ground?

When the sphere is on the ground, its height is zero. We want to know the time when the height of the sphere is zero. That is, we want to know t when h = 0 . We want to know the value of t that makes this equation true:

0 = -16t^{2} + 30t + 78

To solve for t , let's first divide both sides of the equation by 2 .

0 = -8t^{2} + 15t + 39

Now let's solve this quadratic equation using the quadratic formula.

t = \({-15 \pm \sqrt{15^2-4(-8)(39)} \over 2(-8)}\)

t = \({-15 \pm \sqrt{225+1248} \over -16}\)

t = \({-15 \pm \sqrt{1473} \over -16}\)

t = \({-15 + \sqrt{1473} \over -16}\) or t = \({-15 - \sqrt{1473} \over -16}\)

t ≈ -1.461 or t ≈ 3.336

Since t is a measure of time, it should be the positive option.

So...the sphere hits the ground after about 3.336 seconds.

hectictar Jan 3, 2018

#1**+1 **

Just confirming before answer, is this what the question lloks like

Yolanda throws a sphere up in the air from the top of a building. After t^{2} seconds, the height of the sphere above the ground is h^{2} feet, where h=−16t^{2}+30t+78. After how many seconds does the sphere hit the ground?

Rauhan Jan 3, 2018

#2**+1 **

Best Answer

After t seconds, the height of the sphere above the ground is h feet, where

h = -16t^{2} + 30t + 78

After how many seconds does the sphere hit the ground?

When the sphere is on the ground, its height is zero. We want to know the time when the height of the sphere is zero. That is, we want to know t when h = 0 . We want to know the value of t that makes this equation true:

0 = -16t^{2} + 30t + 78

To solve for t , let's first divide both sides of the equation by 2 .

0 = -8t^{2} + 15t + 39

Now let's solve this quadratic equation using the quadratic formula.

t = \({-15 \pm \sqrt{15^2-4(-8)(39)} \over 2(-8)}\)

t = \({-15 \pm \sqrt{225+1248} \over -16}\)

t = \({-15 \pm \sqrt{1473} \over -16}\)

t = \({-15 + \sqrt{1473} \over -16}\) or t = \({-15 - \sqrt{1473} \over -16}\)

t ≈ -1.461 or t ≈ 3.336

Since t is a measure of time, it should be the positive option.

So...the sphere hits the ground after about 3.336 seconds.

hectictar Jan 3, 2018