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.

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opsadnojD · Accepted Answer

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 &amp;amp; c=256 in equation (1), we get h(1)=−1612+256

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

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