You are walking in Paris alongside the Eiffel Tower and suddenly a croissant smacks you on the head and knocks you to the ground. From your handy dandy tourist guidebook you find that the height of the Eiffel Tower is 300.5 m. If you neglect air resistance, calculate how many seconds the croissant dropped before it tagged you on the head.

Answers

Answer 1
Answer: Neglecting air resistance (that is, if you and the Eiffel Tower were in a giant vacuum chamber), the bagel would take 7.804 s to hit you on the head.

I will assume that you are 1.8 m tall, so the distance the bagel falls before it hits you on the head is

300.5 m - 1.8 m = 298.7 m

One of the important equations relating distance and acceleration is

s = v₀ +½at², where

s = distance,
v₀ = initial velocity,
a = acceleration, and
t = time.

I also assume that the person at the top of the tower didn't throw the bagel down at you, so v₀ = 0.

a is the acceleration due to gravity, which is 9.8099 m·s⁻² in Paris.

s = v₀ + ½ at²

298.7 m = 0 + ½ × 9.8099 m·s⁻² × t²

298.7 = 4.9050 t² s⁻²

t² = (298.7 \text m)/( 4.9050 \text s^(-2)) = 60.898 s²

t = \sqrt{60.898 s^(2)} = 7.804 s


Answer 2
Answer:

The time taken for the croissant to hit you on the head assuming you are 1.5 m tall is 7.73 s

Considering the question given above, your height was not stated.

Let your height be 1.5 m

Thus, the height of the Eiffel Tower to your head will be:

Height (h) = (Total height of tower) – (Your assumed height)

Total height of tower = 300.5 m

Your assumed height = 1.5 m

Height (h) = 300.5 – 1.5

Height (h) = 299 m

Finally, we shall determine the time taken for the croissant to hit you on the head.

Height (h) = 299 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

H = ½gt²

299 = ½ × 10 × t²

299 = 5 × t²

Divide both side by 5

t^(2)  = (299)/(5)\n\nt^(2)  = 59.8

Take the square root of both side

t = √(59.8)

t = 7.73 s

Therefore, the time taken for the croissant to hit you on the head assuming you are 1.5 m tall is 7.73 s

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Option 2

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Answer:

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