A ball is thrown upward from the top of a

ahgan3j

ahgan3j

Answered question

2021-12-04

A ball is thrown upward from the top of a 240-foot-high
building. The ball is 256 feet above ground level after 11 seconds, and it reaches ground level in 55 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

Charles Randolph

Charles Randolph

Beginner2021-12-05Added 16 answers

Step 1
Given:
A ball is thrown upward from the top of a 240-foot-high building. The ball is 256 feet above ground level after 11 seconds, and it reaches ground level in 55 seconds. The height above ground is a quadratic function of the time after the ball is thrown.
Calculation:
The formula to find the quadratic function is
h(t)=at2+vt+h0,(1)
Here, h0=240
Substitute the above value in equation (1), we get
h(t)=at2+vt+240..(2)
From the given information,
h(11)=256 and h(55)=0
Step 2
Substitute the values in equation (2), we get
h(11)=a(11)2+v(11)+240
256=121a+11v+240
12la+11v+240256=0
12la+11v16=0(3)
h(55)=a(55)2+(55)+240
0=3025a+55v+240
3025a+55v+240=0..(4)
Solving equation (3) and (4), we get
a=16121 and v=3211
Substitute the values in equation (2), we get
n(t)=16121t2+3211t+240
Step3
Therefore, the quadratic equation of the given function is h(t)=16121t2+3211t+240

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