Question

Мяч бросают вверх с поверхности Земли со скоростью 5м/с. На какой высоте его скорость станет равной 2м/с?

Answer 1

Mv^2/2=mu^2/2+mgh
h=(v^2-u^2)/2g=(5^2-2^2)/(2*10) м = 1,05 м ~ 1 м

Answer 2

Calculation of the Height at which the Speed of the Ball becomes 2 m/s

To calculate the height at which the speed of the ball becomes 2 m/s, we can use the principle of conservation of mechanical energy. According to this principle, the total mechanical energy of the ball remains constant throughout its motion.

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

Total Mechanical Energy (E) = KE + PE

At the initial point when the ball is thrown upwards, its speed is 5 m/s. Let's assume the initial height is h0 and the final height is h.

At the initial point, the total mechanical energy is given by:

E0 = KE0 + PE0

Since the ball is at rest at the initial point, its kinetic energy is zero:

KE0 = 0

The potential energy at the initial point is given by:

PE0 = m * g * h0 where m is the mass of the ball and g is the acceleration due to gravity.

At the final point when the ball reaches a height h, the total mechanical energy is given by:

E = KE + PE

The kinetic energy at the final point is given by:

KE = (1/2) * m * v^2

where v is the speed of the ball.

The potential energy at the final point is given by:

PE = m * g * h According to the principle of conservation of mechanical energy, the total mechanical energy at the initial point is equal to the total mechanical energy at the final point:

E0 = E

Substituting the expressions for KE0, PE0, KE, and PE, we get:

0 + m * g * h0 = (1/2) * m * v^2 + m * g * h

Simplifying the equation, we have:

m * g * h0 = (1/2) * m * v^2 + m * g * h

We can cancel out the mass (m) and acceleration due to gravity (g) from both sides of the equation:

h0 = (1/2) * v^2 + h

Now, we can substitute the given values into the equation to find the height at which the speed of the ball becomes 2 m/s:

h0 = (1/2) * (2 m/s)^2 + h

Simplifying further:

h0 = 2 m^2/s^2 + h

Therefore, the height at which the speed of the ball becomes 2 m/s is h0 + 2 m^2/s^2.

Please note that the above calculation assumes no air resistance and neglects any other external factors that may affect the motion of the ball.

Conclusion

The height at which the speed of the ball becomes 2 m/s is h0 + 2 m^2/s^2.

Let me know if there's anything else I can help you with!