Question

Снаряд Н%3D 15,9 м на две одинаковые части. Через время t %3D 3,00 с после взрыва одна часть падает на Землю под тем местом, где произошел взрыв. С какой разрывается верхней точке траектории на высоте скоростью и под каким углом к горизонту начала двигаться вторая снаряда после взрыва, если первая упала на расстоянии ! выстрела? Сопротивление воздуха не учитывать. 636 м от места

Answer 1

Ответ:

там н#%50.30

Объяснение:

..........................

Answer 2

Problem Analysis

We are given that a projectile, denoted as N, breaks into two equal parts after a certain time t = 3.00 s. One part falls to the ground at the location of the explosion, while the other part continues its trajectory. We need to determine the velocity and angle at which the second projectile starts moving after the explosion, assuming no air resistance.

Solution

To solve this problem, we can use the principles of projectile motion. We'll start by analyzing the motion of the first part of the projectile that falls to the ground.

Let's denote the initial velocity of the projectile as v0 and the angle at which it was launched as θ. We'll assume that the projectile was launched from the ground, so its initial height is zero.

Using the equations of motion, we can find the time it takes for the first part of the projectile to reach the ground. The equation for the vertical displacement of a projectile is given by:

y = v0 * sin(θ) * t - (1/2) * g * t^2

Since the initial height is zero, we can set y = 0 and solve for t:

0 = v0 * sin(θ) * t - (1/2) * g * t^2

Simplifying the equation, we get:

(1/2) * g * t^2 = v0 * sin(θ) * t

Dividing both sides by t and rearranging the equation, we get:

t = (2 * v0 * sin(θ)) / g

Now, we know that the time it takes for the first part of the projectile to reach the ground is t = 3.00 s. We can use this information to find the initial velocity v0 in terms of θ:

3.00 = (2 * v0 * sin(θ)) / g

Simplifying the equation, we get:

v0 = (3.00 * g) / (2 * sin(θ))

Now that we have the initial velocity v0, we can find the horizontal component of the velocity, denoted as vx. The horizontal component of the velocity remains constant throughout the motion of the projectile. We can find vx using the equation:

vx = v0 * cos(θ)

Finally, we can find the vertical component of the velocity, denoted as vy, using the equation:

vy = v0 * sin(θ)

Now, let's move on to the second part of the projectile. Since the two parts of the projectile have the same initial velocity and angle, the second part will also have an initial velocity of v0 and an angle of θ.

To find the velocity and angle at which the second part starts moving after the explosion, we need to consider the motion of the first part and subtract its velocity components from the initial velocity of the second part.

The horizontal component of the velocity of the second part, denoted as vx2, will be equal to the horizontal component of the initial velocity, vx.

The vertical component of the velocity of the second part, denoted as vy2, will be equal to the vertical component of the initial velocity, vy, minus the vertical component of the velocity of the first part, vy1.

Using the equations:

vx2 = vx

vy2 = vy - g * t

Substituting the values of vx and vy, we get:

vx2 = v0 * cos(θ)

vy2 = v0 * sin(θ) - g * t

Now we have the velocity components of the second part of the projectile. We can find the magnitude of the velocity, denoted as v2, using the equation:

v2 = sqrt(vx2^2 + vy2^2)

And we can find the angle at which the second part starts moving, denoted as θ2, using the equation:

θ2 = atan(vy2 / vx2)

Substituting the values of vx2 and vy2, we get:

θ2 = atan((v0 * sin(θ) - g * t) / (v0 * cos(θ)))

Now we have the velocity and angle at which the second part of the projectile starts moving after the explosion.

Answer

The velocity at which the second part of the projectile starts moving after the explosion is v2 = sqrt(vx2^2 + vy2^2), where vx2 = v0 * cos(θ) and vy2 = v0 * sin(θ) - g * t.

The angle at which the second part of the projectile starts moving after the explosion is θ2 = atan((v0 * sin(θ) - g * t) / (v0 * cos(θ))).

Please note that the values of v0, θ, and t need to be substituted into the equations to obtain the final numerical values.

Let's calculate the values using the given information:

Given: - N = 15.9 m (length of the projectile) - t = 3.00 s (time after the explosion) - Distance from the explosion to the location where the first part falls = 636 m

We need to find the velocity and angle at which the second part starts moving after the explosion.

To calculate the values, we need the initial velocity v0 and the angle θ at which the projectile was launched. Unfortunately, these values are not provided in the question. Without these values, we cannot calculate the velocity and angle of the second part of the projectile.

Please provide the values of v0 and θ to proceed with the calculation.