# If an object is thrown vertically upward with an intial 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 ft, t is...

If an object is thrown vertically upward with an intial 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 ft, t is in seconds, and v is in ft/sec)

If a rock is thrown upward from a height of 100 ft, with an initial velocity of 32 ft/sec, solve for the time that it takes for it to hit the ground. (When h = 0)

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given the formula `h=-16t^2+vt+s` ; s=100 (the initial height), v=32 (the initial velocity), and we are asked to solve for t when h=0.

Substituting the known values:

`-16t^2+32t+100=0`

`-4(4t^2-8t-25)=0` Use the quadratic formula on the expression in parantheses:

`t=(8+-sqrt((-8)^2-4(4)(-25)))/(2(4))`

`t=(8+-sqrt(464))/8`

`t=(8+-sqrt(16*29))/8=(8+-4sqrt(29))/8`

`t=1+-sqrt(29)/2`

`t~~-1.69"or"t~~3.69`

The first answer does not makes sense in the context of the problem.

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The ball will hit the ground in `t=1+sqrt(29)/2"sec"` or `t~~3.69"sec"`

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