Question

Пикирующий бомбардировщик движется со скоростью V = 150 км/ч по отрезку прямой, образующей угол α =45° с горизонтальной плоскостью. Цель перемещается по земле со скоростью V2=80км/ч в плоскости движения самолета. На каком расстоянии s по горизонтали от цели должна быть сброшена бомба с высоты Н=4 км, чтобы поразить цель?

Answer 1

1) найдем горизонтальную скорость бомбы

Vx=Vосоs45=150*1,414=212,1 км\ч=Vоy=60м\с

2) найдем время падения бомбы   h=Vоy*t+gt^2\2
4000=60*t+5t^2
800=12t+t^2

t=6+√36+800=30 c

так как t=L\Vотн
L=t*(Voy-Vц)=30(60-22,22)=1733 м - ответ

Answer 2

Problem Analysis

We are given the following information: - The diving bomber is moving with a speed of V = 150 km/h. - The angle between the bomber's path and the horizontal plane is α = 45°. - The target is moving on the ground with a speed of V2 = 80 km/h in the plane of the bomber's motion. - The bomb is dropped from a height of H = 4 km.

We need to find the horizontal distance s from the target at which the bomb should be dropped to hit the target.

Solution

To solve this problem, we can break it down into two components: the horizontal component and the vertical component.

# Horizontal Component

The horizontal component of the bomber's velocity is given by Vx = V * cos(α) Therefore, the horizontal distance covered by the bomber in time t is given by:

s = Vx * t

To find the time t, we need to consider the relative motion between the bomber and the target.

# Relative Motion

The relative velocity between the bomber and the target is the vector difference between their velocities. In this case, the target is moving in the same plane as the bomber, so we only need to consider the horizontal component of the target's velocity.

The horizontal component of the target's velocity is given by V2x = V2. Therefore, the relative velocity between the bomber and the target is:

Vr = Vx - V2x

# Time of Flight

The time of flight t can be calculated by dividing the horizontal distance s by the relative velocity Vr:

t = s / Vr

# Vertical Component

The vertical distance covered by the bomb can be calculated using the equation of motion for vertical motion:

H = (1/2) * g * t^2

where g is the acceleration due to gravity.

# Solving for s

We can substitute the expression for t from the horizontal component equation into the equation for the vertical component:

H = (1/2) * g * (s / Vr)^2

Simplifying the equation, we can solve for s:

s = sqrt(2 * H * Vr^2 / g)

Now we can substitute the given values into the equation to find the horizontal distance s.

Calculation

Given: - V = 150 km/h - α = 45° - V2 = 80 km/h - H = 4 km

First, we need to convert the velocities to m/s and the height to meters: - V = 150 km/h = 150 * (1000/3600) m/s = 41.67 m/s - V2 = 80 km/h = 80 * (1000/3600) m/s = 22.22 m/s - H = 4 km = 4 * 1000 m = 4000 m

Next, we can calculate the horizontal component of the bomber's velocity: - Vx = V * cos(α) = 41.67 * cos(45°) = 29.29 m/s

Then, we can calculate the relative velocity between the bomber and the target: - V2x = V2 = 22.22 m/s - Vr = Vx - V2x = 29.29 - 22.22 = 7.07 m/s

Now, we can calculate the horizontal distance s: - s = sqrt(2 * H * Vr^2 / g) = sqrt(2 * 4000 * 7.07^2 / 9.8) = 282.84 m

Therefore, the bomb should be dropped at a horizontal distance of approximately 282.84 meters from the target to hit it.

Answer

The bomb should be dropped at a horizontal distance of approximately 282.84 meters from the target to hit it.

Note: The calculations provided above are based on the given information and assumptions.