Question

Рыбак , стоящий на крутом берегу реки высотой 3 м, забрасывает блесну под углом 45 к горизонту. Определить начальную скорость блесны , если дальность ее полета в горизонтальном направлении равна 90 м.

Answer 1

1)L=Vo² ·sin2a/g
2)Vo=√Lg/sin2a=√90*10/sin90=√900/1=30 м/с

Answer 2

Problem Analysis

We are given that a fisherman standing on a steep riverbank with a height of 3 m casts a fishing lure at an angle of 45 degrees to the horizontal. The horizontal distance traveled by the lure is 90 m. We need to determine the initial velocity of the lure.

Solution

To solve this problem, we can use the equations of projectile motion. The horizontal and vertical components of the motion are independent of each other.

Let's denote the initial velocity of the lure as V and the angle of projection as θ.

The horizontal distance traveled by the lure is given as 90 m. This is the horizontal component of the displacement. We can use the equation for horizontal displacement:

Horizontal Displacement (x) = V * cos(θ) * t

Since the horizontal displacement is 90 m, we can rewrite the equation as:

90 = V * cos(45) * t

Similarly, the vertical displacement is the height of the riverbank, which is 3 m. We can use the equation for vertical displacement:

Vertical Displacement (y) = V * sin(θ) * t - (1/2) * g * t^2

Since the vertical displacement is 3 m, we can rewrite the equation as:

3 = V * sin(45) * t - (1/2) * 9.8 * t^2

We have two equations with two unknowns (V and t). We can solve these equations simultaneously to find the initial velocity of the lure.

Calculation

Let's solve the equations using the given values:

Horizontal Displacement (x) = 90 m

Vertical Displacement (y) = 3 m

θ = 45 degrees

g = 9.8 m/s^2 (acceleration due to gravity)

Using the equation for horizontal displacement:

90 = V * cos(45) * t

Simplifying the equation:

90 = V * (sqrt(2)/2) * t

t = 90 / (V * (sqrt(2)/2))

Substituting this value of t into the equation for vertical displacement:

3 = V * sin(45) * (90 / (V * (sqrt(2)/2))) - (1/2) * 9.8 * (90 / (V * (sqrt(2)/2)))^2

Simplifying the equation:

3 = (V * (sqrt(2)/2)) * (90 / (V * (sqrt(2)/2))) - (1/2) * 9.8 * (90 / (V * (sqrt(2)/2)))^2

3 = 90 - (1/2) * 9.8 * (90 / V)^2

Simplifying further:

3 = 90 - (1/2) * 9.8 * 8100 / V^2

3 = 90 - 44100 / V^2

44100 / V^2 = 90 - 3

44100 / V^2 = 87

V^2 = 44100 / 87

V^2 = 506.8966

Taking the square root of both sides:

V = sqrt(506.8966)

V ≈ 22.52 m/s

Therefore, the initial velocity of the lure is approximately 22.52 m/s.

Answer

The initial velocity of the fishing lure, when cast by the fisherman standing on a steep riverbank with a height of 3 m at an angle of 45 degrees to the horizontal, is approximately 22.52 m/s.