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

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

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\\eq0`.

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) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\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) = (-(-90) + √(420))/(2*16) = 3.5`

`t_(2) = (-(-90) - √(420))/(2*16) = 2.2`

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