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VikaD [51]
3 years ago
14

A pelican flying along a horizontal path drops a fish from a height of 5.4 m. The fish travels 8.0 m horizontally before it hits

the water below. The acceleration of gravity is 9.81 m/s2 .
a) What was the pelican’s initial speed?
b) If the pelican was traveling at the same speed but was only 2.7 m above the water, how far would the fish travel horizontally before hitting the water below?
Physics
1 answer:
Schach [20]3 years ago
6 0

Answer:

a) V = 6.25 m/s

b) d = 4.6 m

Explanation:

a) <u>The pelican's initial speed is equal to the horizontal speed of the fish</u>. So we can calculate Vx of the fish knowing the fact that it travelled 8.0 m, using the formula for an <em>Uniformly Accelerated Motion</em> (because the fish is freefalling) to calculate the time the fish was falling:

D(t) = 0.5*a*t² + V₀*t + e₀

In this case, V₀ and e₀ are zero, a is gravity's acceleration and D(t) is 8.0 m

8.0 m = 0.5 * 9.81m/s² * t²

t² = 1.63 s²

t = 1.28 s

Thus Vx of the fish is:

8.0 m / 1.28 s = 6.25 m/s

<u>And that's the same initial speed of the pelican.</u>

b) The pelican is traveling at the same speed, so Vx of the fish remains the same, 6.25 m/s. First we calculate the time again:

D(t) = 0.5*a*t²

2.7 m = 0.5 * 9.81m/s² * t²

t² = 0.55 s²

t = 0.74 s

Now we use the formula V = d/t and solve for t:

6.25 m/s = d / 0.74 s

d = 4.6m

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