The body is thrown at an angle to the horizon at a speed of 10 m / s. Determine the speed of the body at a height of 3 meters.
v0 = 10 m / s is the initial velocity with which the body was thrown;
H = 3 meters – body height;
g = 10 m / s ^ 2 – acceleration of gravity.
It is required to determine v1 (m / s) – the speed of the body at height H.
According to the law of conservation of energy, the kinetic energy of the body at the moment of flight will be equal to the sum of the kinetic and potential energies at the height H, that is:
m * v0 ^ 2/2 = m * g * H + m * v1 ^ 2/2, where m is body weight;
v0 ^ 2/2 = g * H + v1 ^ 2/2;
v0 ^ 2 = 2 * g * H + v1 ^ 2;
v1 ^ 2 = v0 ^ 2 – 2 * g * H;
v1 = (v0 ^ 2 – 2 * g * H) ^ 0.5 = (10 ^ 2 – 2 * 10 * 3) ^ 0.5 = (100 – 60) ^ 0.5 =
= 40 ^ 0.5 = 6.3 m / s.
Answer: the speed of the body at a height of 3 meters will be equal to 6.3 m / s.
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