Aball is thrown upward from the top of a 48

Virginia Caron

Virginia Caron

Answered question

2021-12-05

Aball is thrown upward from the top of a 48 foot-high-building. The ball is 64 feet above ground level after 1 second, and it reaches ground level in 3 seconds. The height above ground is a quadratic function of the time after the ball is thrown. Write the equation of this function.

Answer & Explanation

Edward Belanger

Edward Belanger

Beginner2021-12-06Added 11 answers

the height above ground is a quadratic function of the time, t is given by equation
h(t)=at2+bt+c
when t=0,h(0)=48 given
plug given condition
h(0)=a×02+b×0+c
48=c
With gravity, these are usually 16t2 functions, because gravity is 32ftsec2, and the formula has a half in it.
So equation for h(t) will be
h(0)=16t2+bt+48
Solve for the value of b, using given condition
h(t)=16t2+bt+48
when t=3, ball reaches ground so height will be 0
so, h(3)=0
16t2bt+48=0plugt=3
144+3b+48=0
96+3b=0
3b=96
b=32
Final equation for height function will be
h(t)=16t2+32t+48

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