Solve Quadratics: Find Time for Rock Dropped from Washington Monument

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To determine how long it takes for a rock dropped from the 555-foot tall Washington Monument to hit the ground, the correct initial velocity (vo) should be 0, not -9.8. The equation used is h(t) = -16t^2 + vot + ho, where the acceleration due to gravity is accounted for in the formula. The confusion arose from mixing up velocity and acceleration, as the correct acceleration is 32 ft/sec², leading to the -16 in the equation. Completing the square method was attempted, but the initial conditions need to be corrected for accurate time calculation. Properly applying the formula will yield the time it takes for the rock to reach the ground.
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Homework Statement


A rock is dropped from the Washington Monument which is 555 feet tall. Using the formula h(t) = -16t^2 + vot + ho , find approximately how long it will take for the rock to hit the ground.


Homework Equations


h(t) = -16t^2 + vot + ho


The Attempt at a Solution


I'm doing the complete the square method
0 = -16t^2 - 9.8t + 555
-16t^2-9.8t=-555
-16t^2-9.8t +24.01 = -555-384.16 = -939.16
-16(t-4.9)^2 = -939.16
Divide both sides by 16
(t-4.9)^2 = 58.69
Take the square roots
t-4.9 = sqrt58.69

At this point I'm confused as to how to get a number for the time. If anyone can help it'd be a great! thanks.
 
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jubbly said:

Homework Statement


A rock is dropped from the Washington Monument which is 555 feet tall. Using the formula h(t) = -16t^2 + vot + ho , find approximately how long it will take for the rock to hit the ground.


Homework Equations


h(t) = -16t^2 + vot + ho


The Attempt at a Solution


I'm doing the complete the square method
0 = -16t^2 - 9.8t + 555
-16t^2-9.8t=-555
-16t^2-9.8t +24.01 = -555-384.16 = -939.16
-16(t-4.9)^2 = -939.16
Divide both sides by 16
(t-4.9)^2 = 58.69
Take the square roots
t-4.9 = sqrt58.69

At this point I'm confused as to how to get a number for the time. If anyone can help it'd be a great! thanks.
You erroneously put in vo = -9.8, when in fact vo = 0. I think you got velocity (v) and acceleration (g) mixed up. (Incidentally, if you were looking at acceleration, in the ft/sec/sec units, g is 32 ft/sec/sec. That's where the 16 comes from, it was built into the equation already, per the 1/2at^2 term of the kinematic equation).
 
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