Question

A rugby player kicks a rugby ball 1 foot above the ground with an initial vertical velocity of 55 feet per second. The function h=-16t^2+55t+1 represents the height of the rugby ball after t seconds. Use a graphing calculator to estimate when the height of the rugby ball is 45 feet

Answer 1

The height of the rugby ball is 45 feet at t = 1.27s and t = 2.17s

Explanation:

The height of the ball, in feet, after t seconds, is given by the following equation:

`h(t) = -16t^(2) + 55t + 1`

When the height of the rugby ball is 45 feet

This is t for which: h(t) = 45

So

`h(t) = -16t^(2) + 55t + 1`

`45 = -16t^(2) + 55t + 1`

`-16t^(2) + 55t - 44 = 0`

Solving a quadratic equation:

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 = a(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 question:

`-16t^(2) + 55t - 44 = 0`

So
`a = -16, b = 55, c = -44`

`\bigtriangleup = 55^(2) - 4*(-16)*(-44) = 209`

`t_(1) = (-55 + √(209))/(2*(-16)) = 1.27`

`t_(2) = (-55 - √(209))/(2*(-16)) = 2.17`

The height of the rugby ball is 45 feet at t = 1.27s and t = 2.17s