Respuesta :
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, [tex]\theta = 42^{\circ}[/tex]
Now,
The time taken to cover the horizontal distance, d:
[tex]t = \frac{x}{u'_{x}}[/tex]
where
[tex]u'_{x} = u'cos\theta[/tex] = horizontal component of velocity
[tex]t = \frac{0.26}{3.7cos42^{\circ}} = 0.0945 s[/tex]
The maximum height attained by the fish is given by;
[tex]H = u'_{y}t - \frac{1}{2}gt^{2}[/tex]
where
[tex]u'_{y} = u'sin\theta[/tex] = vertical component of velocity
g = acceleration due to gravity
[tex]H = 3.7sin42^{\circ}\times 0.0945 - \frac{1}{2}\times 9.8\times (0.0945)^{2}[/tex]
H = 0.1902 m = 19.02 cm
Answer:
y = 18.95 cm
Explanation:
speed of the water stream is given as
[tex]v = 3.7 m/s[/tex]
also we know that
[tex]\theta = 42 degree[/tex]
now we have
[tex]v_x = v cos42[/tex]
[tex]v_x = 3.7 cos42 = 2.75 m/s[/tex]
also we know that
[tex]v_y = 3.7 sin42 = 2.47 m/s[/tex]
now we know that the time taken by the stream to reach the position of stream is given as
[tex]t = \frac{x}{v_x}[/tex]
[tex]t = \frac{0.26}{2.75}[/tex]
[tex]t = 0.0945 s[/tex]
now the height of the water stream is given as
[tex]y = v_y t + \frac{1}{2}at^2[/tex]
[tex]y = 2.47(0.0945) - \frac{1}{2}(9.81)(0.0945)^2[/tex]
[tex]y = 18.95 cm[/tex]
