A ball is thrown upward from the top of a 48-foot-high building.

Yolanda Jorge

Yolanda Jorge

Answered question

2021-11-28

A ball 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

Susan Yang

Susan Yang

Beginner2021-11-29Added 20 answers

Step 1
Assume that the height of the ball be h ft above the ground after t second.
Then, a quadratic function for the height of the ball will be: h(t)=at2+bt+c.
Since the height of the ball is 64 ft at t=0, therefore
48=a(0)2+b(0)+c
c=48
Since the height of the ball is 64 ft at t=1, therefore
64=a(1)2+b(1)+48
a+b=16
b=a16
Step 2
Since the height of the ball is 0 ft at t=3, therefore
0=a(3)2+b(3)+48
9a+3b=48
3a+b=16
3a+16a=16
2a=32
a=16
And,
b=16(16)
=32
Therefore, the required function is h(t)=16t2+32t+48.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?