Question

An arrow is shot vertically upward at a rate of 170 feet per second from ground level.

*Look at photo attached for the complete question...

Answer 1

t1= 1.1475 sec

t2=9.48 sec

Explanation:

This question asking for the time when an object reach a specific height. I will explain what the formula stand for to make this easier.

h= height = 174feet

v0= initial velocity = 170 feet/s

t= time

h0= initial height= ground level = 0

The question have predetermined formula for the height, so we should use it instead of formula that involve gravity value.

h= -16
`t^(2)` +v0t + h0t

174 feet = -16
`t^(2)` + 170 feet/s x t + 0 x t

-16
`t^(2)` + 170 feet/s x t - 174 feet = 0

8
`t^(2)` - 85 feet/s x t + 87 feet = 0

This equation will be hard to solve since its value in fraction, so lets use the quadratic formula

`x=  (-b±√( b^2-4ac))/(2a)`

`t=  ((85)±√( -85^2-4(8)(87)))/(2(8))`

`t=  (85±\ 66.64)/(16)`

t1=
`(85-\ 66.64)/(16)`

t1= 1.1475

t2=
`(85+\ 66.64)/(16)`

t2=9.48