Question

A ball is thrown downward from the top of a 240​-foot building with an initial velocity of 12 feet per second. The height of the ball h in feet after t seconds is given by the equation h=-16t^2 - 12t + 240. How long after the ball is thrown will it strike the​ ground?

Answer 1

To find when the ball will strike the ground, the quadratic equation h=-16t^2 - 12t + 240 is set to zero and solved for t using the quadratic formula, yielding approximately 3.53 seconds as the time it will take for the ball to hit the ground.

Step-by-step explanation:

To determine how long after the ball is thrown it will strike the ground, we need to find when the height h is equal to 0. This can be found by solving the quadratic equation given by h=-16t2 - 12t + 240. To find the value of t when the ball hits the ground, we set h to 0 and solve the resulting equation.

0 = -16t2 - 12t + 240

To solve this quadratic equation, we can use the quadratic formula:

t = −b ± √(b2 - 4ac) / (2a)

Where a = -16, b = -12, and c = 240. Plugging these values into the quadratic formula, we get:

t = −(-12) ± √((-12)2 - 4(×-16)×240) / (2×-16)

This simplifies to:

t = 12 ± √(144 + 15360) / -32

t = 12 ± √15504 / -32
t = 12 ± 124.514 / -32

Since time cannot be negative, we take the positive root which gives us:

t = (12 + 124.514) / -32

t = 136.514 / -32

t = -4.266

However, since time cannot be negative, we realize that we must take 12 minus the square root:

t = (12 - 124.514) / -32

t = 3.53 seconds (approx.)

Therefore, the ball will strike the ground approximately 3.53 seconds after it is thrown.

Answer 2

`16t^2 +12 t -240 =0`

And we can use the quadratic formula given by:

`t = (-b \pm √(b^2 -4ac))/(2a)`

Where
`a = 16, b = 12, c =-240`. And replacing we got:

`t = (-12 \pm √((12^2) -4*(16)(-240)))/(2*16)`

And after solve we got:

`t_1 = 3.516 s , t_2 = -4.266s`

Since the time cannot be negative our final solution would be
`t = 3.516 s`

Step-by-step explanation:

We have the following function given:

`h(t) = -16t^2 -12 t +240`

And we want to find how long after the ball is thrown will it strike the​ ground, so we want to find the value of t that makes the equation of h equal to 0

`0 = -16t^2 -12 t +240`

And we can rewrite the expression like this:

`16t^2 +12 t -240 =0`

And we can use the quadratic formula given by:

`t = (-b \pm √(b^2 -4ac))/(2a)`

Where
`a = 16, b = 12, c =-240`. And replacing we got:

`t = (-12 \pm √((12^2) -4*(16)(-240)))/(2*16)`

And after solve we got:

`t_1 = 3.516 s , t_2 = -4.266s`

Since the time cannot be negative our final solution would be
`t = 3.516 s`