An airplane wants to drop a load of fish inot a small lake in the mountains. The plane is flying at its min speed 40 m/s, and is 50 meters above the surface of the lake.
a.) Where relative to the lake should the plane be when the fish are dropped so that they land in the middle of the lake?
b.) How fast are the fish going when they reach the lake?
c.) At what angle do they hit the water?
To answer these questions, we'll need to consider the principles of projectile motion. Let's break it down step by step:
a) To find the horizontal distance the fish will travel, we need to determine the time it takes for them to reach the lake.
First, we need to find the vertical distance the fish will fall in relation to the lake's surface. Since the plane is flying at a constant height of 50 meters above, the vertical distance is simply 50 meters.
Next, we can use the formula for vertical distance (d) traveled by an object in free fall at time (t) with an initial vertical velocity (v0):
d = v0t + (1/2)gt^2
In this case, the initial vertical velocity is 0 m/s because the fish are dropped from rest. The acceleration due to gravity (g) is approximately 9.8 m/s^2. Therefore, our equation becomes:
50 = 0 + (1/2)(9.8)t^2
Simplifying this equation, we get:
t^2 = (2 * 50) / 9.8
Solving for t, we find:
t ≈ 3.19 seconds
Now, since the plane is flying at a constant horizontal speed of 40 m/s, the horizontal distance traveled by the fish can be found by multiplying the time (t) by the horizontal speed:
Horizontal distance = 40 m/s * 3.19 s ≈ 127.6 meters
Therefore, the plane should be positioned approximately 127.6 meters in front of the middle of the lake when the fish are dropped, assuming no air resistance or wind affecting the trajectory.
b) The speed of the fish when they reach the lake can be determined using the formula for horizontal velocity (v) based on the horizontal distance (d) and time (t):
v = d / t
Substituting the values we already know:
v = 127.6 m / 3.19 s ≈ 40 m/s
Hence, the fish reach the lake with a speed of approximately 40 m/s.
c) To find the angle at which the fish hit the water, we can use the formula for the angle of projection (θ) in projectile motion. Since the vertical velocity (v0y) is 0 at the highest point of the trajectory, we can calculate the angle using the horizontal velocity (v0x) and vertical velocity (v0y) at the time of release:
tan(θ) = v0y / v0x
The horizontal velocity (v0x) is equal to the constant speed of the plane, which is 40 m/s.
The vertical velocity (v0y) can be found using the equation for vertical velocity (v) based on the time (t) and acceleration due to gravity (g):
v = v0y + gt
Since the fish are dropped from rest, the initial vertical velocity (v0y) is 0. Therefore, the equation simplifies to:
v = gt
Solving for v0y, we get:
v0y = gt ≈ (9.8 m/s^2)(3.19 s) ≈ 31.262 m/s
Now, substituting the values into the formula for the angle of projection:
tan(θ) = 31.262 m/s / 40 m/s
Taking the inverse tangent of both sides to find the angle, we get:
θ ≈ tan^(-1)(0.78155)
Therefore, the fish hit the water at an angle of approximately 38.483 degrees.