A ball is dropped from the top of a 256-foot-high building. The ball is 192 feet

Trent Carpenter

Trent Carpenter

Answered question

2021-10-17

A ball is dropped from the top of a 256-foot-high building. The ball is 192 feet above ground after 2 seconds, 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

opsadnojD

opsadnojD

Skilled2021-10-18Added 95 answers

A ball is dropped from the top of a 256-foot-high building. The ball is 192 feet above ground after 2 seconds, and it reaches ground level in 4 seconds.
Let the height above the ground after t seconds be,
h(t)=at2+bt+c (1)
Since the initial height is 256 foot, therefore h(t)=256 when t=0.
Now put t=0 and h(t)=256 in equation (1), we get
256=a(0)2+b(0)+c
c=256
Since the ball is 192 feet above ground after 2 seconds, therefore h(t)=192 when t=2.
Now put t=2 and h(t)=256 in equation (1), we get
192=a(2)2+b(2)+c
4a+2b+c=192
4a+2b+256=192
4a+2b=64
2a+b=32 (2)
Since the ball reaches the ground level after 4 seconds, therefore h(t)=0 when t=4.
Now put t=4 and h(t)=0 in equation (1), we get
0=a(4)2+b(4)+c
16a+4b+c=0
16a+4b+256=0
16a+4b=256
4a+b=64 (3)
Now subtracting equation (2) from equation (3), we get
(4a+b)(2a+b)=64(32)
2a=32
a=16
Now put a=16 in equation (2), we get
2(16)+b=32
b=0
Now put a=16, b=0 & c=256 in equation (1), we get
h(1)=1612+256

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