Question

The archerfish uses a remarkable method for catching insects sitting on branches or leaves above the waterline. The fish rises to the surface and then shoots out a stream of water precisely aimed to knock the insect off its perch into the water, where the archerfish gobbles it up. Scientists have measured the speed of the water stream exiting the fish's mouth to be 3.7 m/s. An archerfish spots an insect sitting 19 cm above the waterline and a horizontal distance of 26 cm away. The fish aims its stream at an angle of 42 ∘ from the waterline.
Determine the height above the waterline that the stream reaches at the horizontal position of the insect.

Answer 1

Answer:

y = 18.95 cm

Explanation:

speed of the water stream is given as

$v = 3.7 m/s$

also we know that

$\theta = 42 degree$

now we have

$v_x = v cos42$

$v_x = 3.7 cos42 = 2.75 m/s$

also we know that

$v_y = 3.7 sin42 = 2.47 m/s$

now we know that the time taken by the stream to reach the position of stream is given as

$t = \frac{x}{v_x}$

$t = \frac{0.26}{2.75}$

$t = 0.0945 s$

now the height of the water stream is given as

$y = v_y t + \frac{1}{2}at^2$

$y = 2.47(0.0945) - \frac{1}{2}(9.81)(0.0945)^2$

$y = 18.95 cm$

Answer 2

Answer:

The height above the water line that is reached by the stream is 19.02 cm

Solution:

As per the question:

The speed of the stream of water exiting the fish's mouth, u' = 3.7 m/s

The horizontal distance of the distance from the archer fish, x = 26 cm = 0.26 m

The stream aimed by the fish at an angle, $\theta = 42^{\circ}$

Now,

The time taken to cover the horizontal distance, d:

$t = \frac{x}{u'_{x}}$

where

$u'_{x} = u'cos\theta$ = horizontal component of velocity

$t = \frac{0.26}{3.7cos42^{\circ}} = 0.0945 s$

The maximum height attained by the fish is given by;

$H = u'_{y}t - \frac{1}{2}gt^{2}$

where

$u'_{y} = u'sin\theta$ = vertical component of velocity

g = acceleration due to gravity

$H = 3.7sin42^{\circ}\times 0.0945 - \frac{1}{2}\times 9.8\times (0.0945)^{2}$

H = 0.1902 m = 19.02 cm