Question

Тело массой 100г брошено вниз с высоты 20м с начальной скоростью 20м/с под углом 30 градусов к горизонту. Чему равна максимальная потенциальная энергия тела после удара о землю, если удар абсолютно упругий? Сопротивление воздуха не учитывать.

Answer 1

100 г = 0,1 кг. V = 30 м/с - 10 м/с^2 * 2 с = 10 м/с. E(кин.) = mv^2/2 = 0,1 кг * (10 м/с)^2/2 = 5 Дж. Е(потенц.) = mgh. h = gt^2/2 = 10 м/с^2 * (2 с)^2/2 = 20 м. Е(потенц.) = 0,1 кг * 10 м/с^2 * 20 м = 20 Дм

Answer 2

Calculation of Maximum Potential Energy

To calculate the maximum potential energy of the object after it hits the ground in an absolutely elastic collision, we need to consider the initial kinetic energy and the change in potential energy.

Given: - Mass of the object (m) = 100 g = 0.1 kg - Initial height (h) = 20 m - Initial velocity (v0) = 20 m/s - Launch angle (θ) = 30 degrees

First, let's calculate the initial kinetic energy (KE) of the object using the formula:

KE = (1/2) * m * v0^2 .

Substituting the given values:

KE = (1/2) * 0.1 kg * (20 m/s)^2 = 20 J

Next, let's calculate the maximum potential energy (PE) of the object after it hits the ground. The maximum potential energy occurs when the object is at its highest point, which is the initial height (h).

PE = m * g * h .

Where: - g is the acceleration due to gravity, which is approximately 9.8 m/s^2.

Substituting the given values:

PE = 0.1 kg * 9.8 m/s^2 * 20 m = 19.6 J

Therefore, the maximum potential energy of the object after it hits the ground in an absolutely elastic collision is 19.6 J.

Please note that this calculation assumes no energy loss due to factors such as air resistance or friction.