If an object is thrown vertically upward with an initial velocity of v, from an original position of s, the height h at any time t is given by: h = 16t^2 + vt + s. (Where h and s are in feet, t is in seconds and v is in ft/sec).

A ball is thrown upward with initial velocity of 96 ft/sec from the top of a 100 ft bridge. Determine the time that it takes for the ball to get to a height of 200 ft from the ground.

Round answers to two decimals, separated by a comma.

### 1 Answer | Add Yours

u= 96 ft/sec

s=100 ft (height of the top of bridge)

h= 200 ft (from ground)

200=16 t^2+96t+100

16t^2+96t-100=0

4t^2+24t-25=0

`t=(-24+-sqrt(24^2+4xx4xx25))/16`

`=(-24+-31.241)/16`

`t>0 therefore`

`t=7.241/16`

`=.453` sec.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes