rozana,a fisherwomen throws her net in a lake.the net falls in the lake when it is 2 meters above the ground at 5 meters per second at angle of 60 degrees.(assume that the acceleration due to gravity is -9.8m/s^2).find the vector function of the net at the starting position and find maximum height rozana's net can reach and the time it reached that height.
To find the vector function of the net at the starting position, we need to break down the initial velocity into its horizontal and vertical components.
The horizontal component can be found using the formula: Vx = V * cos(θ)
where V is the magnitude of the velocity (5 m/s) and θ is the angle of projection (60 degrees).
Vx = 5 * cos(60) = 2.5 m/s
The vertical component can be found using the formula: Vy = V * sin(θ)
where V is the magnitude of the velocity (5 m/s) and θ is the angle of projection (60 degrees).
Vy = 5 * sin(60) = 4.33 m/s
Now, let's derive the equations for the horizontal and vertical position of the net as a function of time.
Horizontal motion:
The horizontal velocity (Vx) remains constant throughout, and there is no horizontal acceleration. Therefore, the horizontal position (x) is given by: x = Vx * t
Vertical motion:
The net starts at a height of 2 meters above the ground, with an initial vertical velocity (Vy). The acceleration due to gravity (-9.8 m/s²) acts in the downward direction.
Using the second equation of motion: h = V0y * t + (1/2) * a * t²
where h is the vertical displacement, V0y is the initial vertical velocity, a is the acceleration due to gravity, and t is the time.
Taking the downward direction as negative:
h = 2 + (4.33 * t) - (4.9 * t²)
Therefore, the vector function representing the position of the net as a function of time can be given as:
r(t) = [x, y] = [Vx * t, 2 + (4.33 * t) - (4.9 * t²)]
To find the maximum height reached by the net, we need to determine the time (t) at which the vertical velocity becomes zero. This happens when the net reaches its peak and starts falling back down.
Using the equation for vertical velocity: Vy = V0y + (a * t)
where Vy is the vertical velocity, V0y is the initial vertical velocity, a is the acceleration due to gravity, and t is the time.
Since Vy = 0 at the maximum height, we can solve for t:
0 = 4.33 + (-9.8 * t)
Solving the equation, we get:
t = 4.33 / 9.8 ≈ 0.441 seconds
Substituting this value of t into the equation for vertical position, we can find the maximum height:
h = 2 + (4.33 * 0.441) - (4.9 * (0.441)²
h ≈ 3.78 meters
So, the maximum height reached by the net is approximately 3.78 meters, and it takes approximately 0.441 seconds to reach that height.