Question

A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is 1. After how many seconds does the ball strike the ground?

Answer 1

`t=6.96s`

Step-by-step explanation:

From this exercise, our knowable variables are hight and initial velocity

`v_(oy)=96ft/s`

`y_(o)=112ft`

To find how much time does the ball strike the ground, we need to know that the final position of the ball is y=0ft

`y=y_(o)+v_(oy)t+(1)/(2)gt^(2)`

`0=112ft+(96ft/s)t-(1)/(2)(32.2ft/s^(2))t^(2)`

Solving for t using quadratic formula

`t=\frac{-b±\sqrt{b^(2)-4ac } }{2a}`

`a=-(1)/(2) (32.2)\\b=96\\c=112`

`t=-0.999s` or
`t=6.96s`

Since time can't be negative the answer is t=6.96s