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forsale [732]
3 years ago
9

A pelican flying along a horizontal path drops a fish from a height of 4.7 m. The fish travels 9.3 m horizontally before it hits

the water below. What was the pelican’s initial speed? The acceleration of gravity is 9.81 m/s 2 . Answer in units of m/s. If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?
Physics
1 answer:
slavikrds [6]3 years ago
8 0

Answer:

(A) 9.5 m/s

(B) 5.225 m

Explanation:

vertical height (h) = 4.7 m

horizontal distance (d) = 9.3 m

acceleration due to gravity (g) = 9.8 m/s^{2}

initial speed of the fish (u) = 0 m/s

(A) what is the pelicans initial speed ?

  • lets first calculate the time it took the fish to fall

s = ut + (\frac{1}{2}) at^{2}

since u = 0

s =  (\frac{1}{2}) at^{2}

t = \sqrt{\frac{2s}{a} }

where a = acceleration due to gravity and s = vertical height

t = \sqrt{\frac{2 x 4.7 }{9.8} } = 0.98 s

  • pelicans initial speed = speed of the fish

speed of the fish = distance / time = 9.3 / 0.98 = 9.5 m/s

initial speed of the pelican = 9.5 m/s

(B) If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?

vertical height = 1.5 m

pelican's speed = 9.5 m/s

  • lets also calculate the time it will take the fish to fall

s = ut + (\frac{1}{2}) at^{2}

since u = 0

s =  (\frac{1}{2}) at^{2}

t = \sqrt{\frac{2s}{a} }

where a = acceleration due to gravity and s = vertical height

t = \sqrt{\frac{2 x 1.5 }{9.8} } = 0.55 s

 

distance traveled by the fish = speed x time = 9.5 x 0.55 = 5.225 m

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θ  = 12.95º

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     θ  = cos⁻¹-1 ((0.20 - 0.00509) /0.20)

      θ  = 12.95º

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