Respuesta :
Answer:11.101 m/s
Explanation:
Given
Velocity of eagle=3.1 m/s
Eagle is at height of 5.8 m
Fish is dropped with a horizontal velocity of 3.1 m/s
time taken by fish to reach lake is
[tex]h=ut+\frac{gt^2}{2}[/tex]
here u=0 (vertical velocity)
[tex]5.8=\frac{9.8\times t^2}{2}[/tex]
t=1.087 s
Vertical velocity acquired by fish is v=u+at
[tex]v=0+9.8\times 1.087=10.66 m/s[/tex]
Horizontal velocity=3.1 m/s
Therefore net velocity[tex]=\sqrt{10.66^2+3.1^2}=11.101 m/s[/tex]
[tex]tan\theta =\frac{v_y}{v_x}=\frac{10.66}{3.1}=3.43[/tex]
[tex]\theta =73.74^{\circ}[/tex] below horizontal
The magnitude of the velocity and the angle is,
- A) The magnitude of the velocity of the fish relative to the water is 11.101 m/s.
- B) The angle, by which the fish's velocity is directed below the horizontal when the fish hits the water is 73.74 degrees.
What is third equation of motion?
The third equation of motion gives the velocity of a object under uniform acceleration.
It can be given as,
[tex]s=ut+\dfrac{1}{2} at^2[/tex]
Here, [tex]u[/tex] is the initial velocity, [tex]t[/tex] is the time, and [tex]a[/tex] is the acceleration.
Given information-
An eagle is flying horizontally at a speed of 3.1 m/s.
The fish falls below in lake at a height of 5.8 m.
- A) Calculate the magnitude of the velocity of the fish relative to the water when it hits the water in m/s.
As the fish is falling by the gravitational force and its initial velocity is zero.
The distance of fish from water is 5.8. Thus put the values in the above equation as,
[tex]5.8=0+\dfrac{1}{2}\times9.81\times t^2\\t=1.087\rm s[/tex]
As the horizontal speed is given in the problem. To find the net velocity we need to find the vertical velocity first. By the first equation of motion,
[tex]v_v=u+at\\v_v=0+9.81\times1.087\\v_v=10.66[/tex]
Thus the net velocity of the body is,
[tex]v_n=\sqrt{v_h^2+v_v^2}\\v_n=\sqrt{3.1^2+10.66^2}\\v_n=11.101\rm m/s[/tex]
Hence the magnitude of the velocity of the fish relative to the water is 11.101 m/s.
- B) The angle by which the fish's velocity is directed below the horizontal when the fish hits the water.
The angle is the arc tan of the ratio of vertical velocity to the horizontal velocity. thus,
[tex]\theta=\tan^- (\dfrac{v_v}{v_h}\\ \theta=\tan^- (\dfrac{10.66}{3.1})\\\theta=73.74^o[/tex]
Hence the angle is 73.74 degrees.
Thus,
- A) The magnitude of the velocity of the fish relative to the water is 11.101 m/s.
- B) The angle, by which the fish's velocity is directed below the horizontal when the fish hits the water is 73.74 degrees.
Learn more about the third equation of motion here;
https://brainly.com/question/8898885
