If an object is thrown vertically upward with an initial velocity of v, from an originial position of s, the height at any given time t is given by:

h = -16t^2 + vt +s (h and s are in ft, t is in seconds, and v is in ft/sec)

A ball is thrown upward with an initial velocity of 96 ft/sec from the top of a 100 foot bridge. Determine the time that it takes for the ball to get to a height of 200 ft. Round answers to 2 decimals, separated by commas.

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The height of the object at time t is given by:

`h=-16t^2+vt+s`

We are given `s=100,v=96` and we are asked to find t when h=200. Substituting the values we get:

`200=-16t^2+96t+100`

`16t^2-96t+100=0`

`4(4t^2-24t+25)=0` Use the quadratic formula on the expression in the parantheses:

`t=(24+-sqrt((-24)^2-4(4)(25)))/(2(4))`

`t=(24+-sqrt(176))/8`

`t=3+-sqrt(11)/2`

The solutions are `t~~1.34"sec"` and `t~~4.66"sec"` ; the first as the ball rises, and then as it falls back down.

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