Solve the problem.
-A person at the top of a 600 foot tall building drops a yellow ball. The height of the yellow ball is given by the equation $$h = - 16 t ^ { 2 } + 600$$ where h is measured in feet and t is the number of seconds since the yellow ball was
Dropped. A second person, in the same building but on a lower floor that is 408 feet from the ground, drops a
White ball 3 seconds after the yellow ball was dropped. The height of the white ball is given by the equation $$h = - 16 ( t - 3 ) ^ { 2 } + 408$$ where h is measured in feet and t is the number of seconds since the yellow ball was
Dropped. Find the time that the balls are the same distance above the ground and find this distance.
A) 2.5 sec; 500 ft
B) 3.5 sec; 404 ft
C) 3 sec; 456 ft
D) 4 sec; 344 ft

Question

Quizplus · Accepted Answer

To find when the balls are at the same height, set the two equations equal to each other and solve for $t$:$$-16t^2 + 600 = -16(t - 3)^2 + 408$$Expanding the right side gives:$$-16t^2 + 600 = -16(t^2 - 6t + 9) + 408$$Simplifying:$$-16t^2 + 600 = -16t^2 + 96t - 144 + 408$$The $-16t^2$ terms cancel out, leaving:$$600 = 96t + 264$$Subtract 264 from both sides:$$336 = 96t$$Divide by 96:$$t = 3.5$$To find the height at $t = 3.5$, substitute back into one of the original equations:$$h = -16(3.5)^2 + 600$$$$h = -16(12.25) + 600$$$$h = -196 + 600$$$$h = 404$$So, at $t = 3.5$ seconds, both balls are 404 feet above the ground.

Solve the Problem h=−16t2+600h = - 16 t ^ { 2 } + 600h=−16t2+600

Solve the Problem $h = - 16 t ^ { 2 } + 600$