Question

8.The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water from the pond surface with an initial speed of 2.30 m/s at an angle of 19.5° above the horizontal. When the stream of water reaches a beetle on a leaf at height above the water’s surface, it is moving horizontally. What is the vector that describes the displacement from the archerfish to the beetle?

Answer 1

At max height, vertical component of velocity is zero.

Answer 2

Given that

Initial speed of archerfish = V0 = 2.30 m/sec at 19.5 deg above the horizontal

V0x = Initial Horizontal velocity = V0*cos = 2.30*cos 19.5 deg = 2.17 m/sec

V0y = Initial Vertical velocity = V0*sin = 2.30*sin 19.5 deg = 0.77 m/sec

Now since there is no acceleration in horizontal direction, So horizontal velocity of fish will remain constant

Also acceleration in vertical direction = g = 9.81 m/sec^2 (In downward direction)

Now given that when water reaches beetle, it's moving horizontally, which means at this point vertical velocity is zero

Using 1st kinematic equation in vertical direction:

Vfy = V0y + a*t

Vfy = 0 m/sec, at beetle

a = -g = -9.81 m/sec (-ve sign since g is downward)

So,

t = (Vfy - V0y)/a = (0 - 0.77)/(-9.81) = 0.0785 sec

Now in this time vertical distance traveled by water will be:

Using 2nd kinematic equation:

y = V0y*t + (1/2)*a*t^2

y = 0.77*0.0785 + (1/2)*(-9.81)*(0.0785)^2

y = 0.0302 m

In the same time horizontal distance traveled by water will be:

Range = x = Speed*time (since there is no acceleration horizontal direction)

x = 2.17*0.0785 = 0.1703 m

Suppose initial position of fish is origin (0, 0), then displacement vector will be:

d = (xf - xi) i + (yf - yi)

f = final position & i = initial position

d = (0.1703 - 0) i + (0.0302 - 0) j

d = (0.1703 i + 0.0302 j) m

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