Respuesta :
Using the kinematic equations below, we can calculate the initial velocity required.
Angle of projectile = 60 degrees
Acceleration due to gravity (Ay) = -10 m/s^2 (negative because downward)
Height of projectile (Dy) = 2m
Vfy^2=Voy^2 +2*Ay*Dy
Vfy = 0 m/s because the vertical velocity slows to zero at the height of its trajection.
So... 0 = Voy^2 + 2(-10)(2)
0 = Voy^2 - 40
40 = Voy^2
Sqrt40 = Voy
6.32 m/s = Voy
THIS IS NOT THE ANSWER. THIS IS JUST THE INITIAL VELOCITY IN THE Y DIRECTION.
Using trigonometry, Tan 60 = Voy/Vox. Tan 60 = 6.32/Vox. Vox*Tan 60 = Vox
Vox = 10.95 m/s. Now, using Vox = 10.95 and Voy = 6.32, we can use pythagorean theorem to find the total Vo. A^2 +B^2 = C^2
10.95^2 + 6.32^2 = C^2
Solving for C = 12.64 m/s
This is the velocity required to hit the surface. You can also calculate a bunch of other stuff now using the other kinematic equations.
V = 12.64 m/s
Answer: To hit the target fish must spit with the velocity of 7.30 m/s.
Explanation:
Highest point of the trajectory = H = 2 m
Angle at which archer fish spit = θ = 60°
Acceleration due to gravity = g = 10 [tex]m/s^2[/tex]
The maximum height of the projectile is given as:
[tex]H=\frac{u^2\sin^2\theta }{2g}[/tex]
[tex]2 m=\frac{u^2\sin^2(60^o)}{2\times 10 m/s^2}[/tex]
[tex]\sin 60^o=\frac{\sqrt{3}}{2}[/tex]
[tex]u^2=\frac{2 m\times 2\times 10 m/s^2\times 4}{3}=53 m^2/s^2[/tex]
[tex]u=7.30 m/s[/tex]
To hit the target fish must spit with the velocity of 7.30 m/s.
