Question

Маленький стальной шарик свободно падает из состояния покоя с высоты 50 метров найти скорость удара шарика о землю сопротивлением воздуха принебречь

Answer 1

Потенциальная энергия шарика mgh   кинетическая при ударе о землю
mv²/2  →  mgh = mv²/2   v²=2gh   v = √2gh
v=√2*10*50=√1000=10√10≈ 31.6 м/сек

Answer 2

Calculation of the Impact Velocity of the Steel Ball

To find the impact velocity of the steel ball when it hits the ground, we can use the principle of conservation of energy. The potential energy of the ball at the initial height is equal to the kinetic energy of the ball just before impact.

The potential energy (PE) of an object at a certain height is given by the formula:

PE = mgh

Where: - m is the mass of the object (assumed to be constant) - g is the acceleration due to gravity (approximately 9.8 m/s^2) - h is the height from which the object falls

The kinetic energy (KE) of an object is given by the formula:

KE = (1/2)mv^2

Where: - m is the mass of the object (assumed to be constant) - v is the velocity of the object

Since the potential energy at the initial height is equal to the kinetic energy just before impact, we can equate the two equations:

PE = KE

mgh = (1/2)mv^2

Simplifying the equation, we can cancel out the mass m:

gh = (1/2)v^2

Solving for v, the impact velocity, we get:

v = sqrt(2gh)

Now we can substitute the given values into the equation to find the impact velocity.

Given: - Height (h) = 50 meters - Acceleration due to gravity (g) = 9.8 m/s^2

Substituting these values into the equation, we get:

v = sqrt(2 * 9.8 * 50)

Calculating this expression, we find that the impact velocity of the steel ball is approximately 31.3 m/s.