A ball is thrown upward from the top of a 128-foot-high building. The

kdgg0909gn

kdgg0909gn

Answered question

2021-12-05

A ball is thrown upward from the top of a 128-foot-high building. The ball is 144 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 the equation of this function.

Answer & Explanation

Leory2000

Leory2000

Beginner2021-12-06Added 10 answers

Step1
Given data:
The initial height of the ball is hi=128m.
The height of the ball after one second is h1=144m.
The height of the ball after four seconds is hf=0.
The expression for the height of the ball is,
y=ax2+bx+c
Here, x is time after the ball is thrown.
Step2
Apply the first condition in the above expression, substitute 0 for x, and 128 for y.
128=a(0)2+b(0)+c
c=128
The expression for height is,
y=ax2+bx+128
Apply the second condition in the above expression, substitute 1 for x, and 144 for y.
144=a(1)2+b(1)+128
a+b=16
b=16a
Apply the third given condition in the expression for the balls height, substitute 4 for x, and 0 for y.
0=a(4)2+b(4)+128
16a+4b=128
4a+b=32
Step3
Substitute (16-a) for b in the above equation.
4a+16a=32
3a=48
a=16
The value of bis,
b=16(16)
=32
Substitute the values of the constant in the expression of the height of the ball.
y=l6x2+32x+128
Thus, the height of the ball at any time x can be express as y=16x2+32x+128.

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