Question

A ball is thrown vertically in the air with a velocity of 90ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in seconds, the ball is at a height of 120ft. Round your answer(s) to the nearest tenth of a second.

Answer 1

Answer:

The ball is at a height of 120ft when t = 2.2 and when t = 3.5.

Step-by-step explanation:

$h = -16t^{2} + v(0)t$

A ball is thrown vertically in the air with a velocity of 90ft/s.

This means that v(0) = 90. So

$h = -16t^{2} + 90t$

Use the projectile formula h=−16t2+v0t to determine at what time(s), in seconds, the ball is at a height of 120ft.

This is t when

$-16t^{2} + 90t = 120$

$16t^{2} - 90t + 120 = 0$

Finding t

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this problem:

$16t^{2} - 90t + 120 = 0$

So $a = 16, b = -90, c = 120$

$\bigtriangleup = (-90)^{2} - 4*16*120 = 420$

$t_{1} = \frac{-(-90) + \sqrt{420}}{2*16} = 3.5$

$t_{2} = \frac{-(-90) - \sqrt{420}}{2*16} = 2.2$

The ball is at a height of 120ft when t = 2.2 and when t = 3.5.

Answer 2

Answer:

$t=1.7 \ sec\ , t=4.0\ sec$

Step-by-step explanation:

Vertical Throw

It refers to a situation where an object is thrown verticaly upwards with some inicial speed v_o and let in free air (no friction) until it completes its movement up and finally returns to the very same point of lauch. The only acting force is gravity

The projectile formula is given as

$h=-16t^2+v_0t$

where t is time in seconds, h is the height in feet and v is the speed in ft/sec

We are required to find the time t where h=120 ft, knowing $v_o=90\ ft/sec$

$-16t^2+90t=120$

Rearranging

$-16t^2+90t-120=0$

This is a second-degree equation which will be solved with the formula

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle x=\frac{-90\pm \sqrt{90^2-4(-16)(-120)}}{2(-16)}$

$\displaystyle x=\frac{-90\pm \sqrt{1380}}{-32}$

Two solutions are obtained

$\boxed{t=1.7 \ sec\ , t= 4.\ sec}$

Both solutions are possible because the ball actually is at 120 ft in its way up and then when going down