Question

СРОЧНО. ДАМ 35 БАЛЛОВ. высоты 1 м вертикально вниз бросили шарик с некоторой скоростью. После удара о землю шарик поднялся на высоту 3м с какой скоростью бросали вниз шарик если в процессе удара 50% механической энергии шарика потерялось. сопротивлением воздуха пренебречь. ​

Answer 1

h₁=1м

h₂=3м

E₂=0.5E₁

v-?

0.5E₁=E₂

0.5(E(k)+E(n1))=E(n2)

0.5(mv²/2 + mgh₁)=mgh₂

0.5(v²/2+gh₁)=gh₂

v²/2 + gh₁=2gh₂

v²/2=2gh₂-gh₁

v²=2g(2h₂-h₁)

v=√(4g(h₂-h₁))=√(2×10 м/с²×(2×3м-1м))=√(100м²/с²)=10м/с

Ответ:10м/с

Answer 2

Calculation of the Initial Velocity of the Ball

To calculate the initial velocity at which the ball was thrown downwards, we can use the principle of conservation of mechanical energy. According to the problem statement, the ball loses 50% of its mechanical energy during the impact with the ground. We can assume that the ball reaches its maximum height of 3m after bouncing back up.

The mechanical energy of the ball can be expressed as the sum of its kinetic energy (KE) and potential energy (PE):

Initial mechanical energy (Ei) = Final mechanical energy (Ef)

The initial mechanical energy of the ball is equal to the sum of its kinetic energy and potential energy before it was thrown downwards:

Ei = KEi + PEi

The final mechanical energy of the ball is equal to the sum of its kinetic energy and potential energy when it reaches its maximum height after bouncing back up:

Ef = KEf + PEf

Since the ball loses 50% of its mechanical energy during the impact, we can express the final kinetic energy as half of the initial kinetic energy:

KEf = 0.5 * KEi

The potential energy at the maximum height is given by:

PEf = m * g * h

where: - m is the mass of the ball (which is not given in the problem statement) - g is the acceleration due to gravity (approximately 9.8 m/s^2) - h is the maximum height reached by the ball (3m)

Substituting the expressions for KEf and PEf into the equation for Ef, we get:

Ef = 0.5 * KEi + m * g * h

Since the ball is thrown vertically downwards, the initial potential energy is zero:

PEi = 0

Therefore, the equation for the initial mechanical energy becomes:

Ei = KEi

Substituting the expressions for Ef and Ei into the equation for the conservation of mechanical energy, we have:

KEi = 0.5 * KEi + m * g * h

Simplifying the equation, we get:

KEi = 2 * m * g * h

The kinetic energy of an object is given by:

KE = 0.5 * m * v^2

where: - m is the mass of the object - v is the velocity of the object

Substituting the expression for KEi into the equation for kinetic energy, we get:

0.5 * m * v^2 = 2 * m * g * h

Simplifying the equation, we find:

v^2 = 4 * g * h

Taking the square root of both sides of the equation, we get:

v = 2 * sqrt(g * h)

Now we can calculate the initial velocity at which the ball was thrown downwards using the given values: - g = 9.8 m/s^2 (acceleration due to gravity) - h = 3 m (maximum height reached by the ball)

Substituting these values into the equation, we find:

v = 2 * sqrt(9.8 * 3) ≈ 8.85 m/s

Therefore, the ball was thrown downwards with an initial velocity of approximately 8.85 m/s.