Question

An object is thrown upward from the top of a 144​-foot building with an initial velocity of 128 feet per second. The height h of the object after t seconds is given by the quadratic equation h= -16t^2 + 128t + 144. After how many seconds will the ball pass the top of the building

Answer 1

The time for the ball to pass the top of the building is determined as 4 s.

Time of motion of the ball

The time for the ball to pass the top of the building is obtained when the velocity of the ball is zero.

h = -16t² + 128t + 144

v = dh/dt

v = -32t + 128

at maximum height, v = 0

0 = -32t + 128

32t = 128

t = 128/32

t = 4 s

Thus, the time for the ball to pass the top of the building is determined as 4 s.

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Answer 2

After 8 seconds, the ball will pass the top of the building if the object is thrown upward from the top of a 144​-foot building with an initial velocity of 128 feet per second.

What is a quadratic equation?

Any equation of the form $\rm ax^2+bx+c=0$  where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

$\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}$

We have:

An object is thrown upward from the top of a 144​-foot building with an initial velocity of 128 feet per second.

When the ball pass the top of the building, h = 144 foot

Plug h = 144 foot in h= -16t² + 128t + 144

144 = -16t² + 128t + 144

$\rm -16t^2+128t=0$

t = 0 or t = 8 seconds

At t = 0, initially, the ball was thrown upward.

At t = 8 seconds, the ball passes the top of the building

Thus, after 8 seconds, the ball will pass the top of the building if the object is thrown upward from the top of a 144​-foot building with an initial velocity of 128 feet per second.

Learn more about quadratic equations here:

brainly.com/question/2263981

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