Question

A ball is launched into the sky at 19.6 feet per second from a 58.8 meter tall building. The equation for the ball’s height, h, at time t seconds is h = -4.9t2 + 19.6t + 58.8 . When will the ball strike the ground?

Answer 1

Therefore,

The ball strike the ground at time , t = 6 s

Explanation:

Given:

The equation for the ball’s height, h, at time t seconds is

`h=-4.9t^(2)+19.6t+58.8`

To Find:

time, t = ? (When the ball strike the ground)

Solution:

When the ball strike the ground height, h will be ZERO,

So put h = 0 in above given equation for time,

`0=-4.9t^(2)+19.6t+58.8`

Which is a Quadratic Equation

Dividing throughout by - 4.9 we get

`0=t^(2)-4t-12`

On Factorizing, splitting the middle term we get

`t^(2)-6t+4t-12=0`

`t(t-6)+2(t-6)=0\\\\(t+2)(t-6)=0\\\\t=-2\ or\ t=6`

As time cannot be negative,

t = 6 s

Therefore,

The ball strike the ground at time , t = 6 s