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.
The height of the object at time t is given by:
We are given `s=100,v=96` and we are asked to find t when h=200. Substituting the values we get:
`4(4t^2-24t+25)=0` Use the quadratic formula on the expression in the parantheses:
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.