Question

Какую скорость при выстреле приобретает пушка массой 1 т, если она стреляет под углом 60 градусов к горизонту? Масса снаряда 100 кг, скорость 300 м/с.

Answer 1

M1V1-m2V2*cos60=0
v1=(m2v2*cos60)\m1=(100*300*0.5)\1000=15м\с

Answer 2

M1v1cosα=m2v2
v2=m1v1cosα/m2=100*300*0.5/1000=15 м/c

Answer 3

Calculation of the Cannonball's Velocity

To calculate the velocity acquired by a cannonball when fired at an angle of 60 degrees to the horizon, we can use the principles of projectile motion. The initial velocity of the cannonball is given as 300 m/s, and its mass is 100 kg. The cannon itself has a mass of 1 ton (1000 kg).

The velocity acquired by the cannonball can be calculated using the following formula:

v = sqrt((2 * m * g * sin(theta)) / (1 + (m_cannon / m)))

Where: - v is the velocity acquired by the cannonball - m is the mass of the cannonball (100 kg) - g is the acceleration due to gravity (approximately 9.8 m/s^2) - theta is the angle of projection (60 degrees) - m_cannon is the mass of the cannon (1000 kg)

Substituting the given values into the formula, we get:

v = sqrt((2 * 100 * 9.8 * sin(60)) / (1 + (1000 / 100)))

Simplifying the equation further:

v = sqrt((1960 * sqrt(3)) / 11)

Evaluating the expression:

v ≈ 240.2 m/s

Therefore, the cannonball will acquire a velocity of approximately 240.2 m/s when fired at an angle of 60 degrees to the horizon.