SOLUTION: a ball is thrown upward from the top of a 64-foot-high building. the ball is 96 feet above ground level after 1 second, and it reaches ground level in 4 seconds. the height above

Question 1000335: a ball is thrown upward from the top of a 64-foot-high building. the ball is 96 feet above ground level after 1 second, and it reaches ground level in 4 seconds. the height above ground is a quadratic function of the time after the ball is thrown. write an equation of this function?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
With gravity, these are usually -16t^2 functions, because gravity is -32 ft/sec^2, and the formula has a half in it. There is force behind the ball, yet unknown, and there is a +64, which is the feet above ground.
-16t^2+ bt+64=0, when t=4
-256+4b+64=0
4b=-192
b=48
This is the equation: f(t)= -16t^2+48t+64.
Check this.
At t=0, the ball is 64 feet off the ground
At 4 seconds, -256+48(4)+64=0
The highest point time value is -b/2a=-48/-32=1.5 seconds.
f(1.5)=-16(2.25)+48(1.5)+64=-36+136=100 feet, the vertex.

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