A body with a mass of 1 kg, thrown at an angle a to the horizon, at a height of 1.4m had a speed of 6m / s
A body with a mass of 1 kg, thrown at an angle a to the horizon, at a height of 1.4m had a speed of 6m / s. The minimum value of the impulse for the entire time of movement is 4 kg * m / s. Find the angle a
Problem data: m (the mass of the thrown body) = 1 kg; h (considered height) = 1.4 m; V (speed at the considered height) = 6 m / s; Pmin (minimum impulse, highest lifting point) = 4 kg * m / s.
Constants: g (acceleration due to gravity) ≈ 10 m / s2.
1) Speed at the highest lifting point: V0x = V0 * cosα = Рmin / m, whence V0 = Рmin / (m * cosα).
2) The law of conservation of energy: m * V0 ^ 2/2 = m * g * h + m * V ^ 2/2.
V0 ^ 2 = 2m * g * h + m * V ^ 2.
(Pmin / (m * cosα)) ^ 2 = 2m * g * h + m * V ^ 2.
(4 / (1 * cosα)) ^ 2 = 2 * 1 * 10 * 1.4 + 1 * 6 ^ 2.
(4 / cosα) ^ 2 = 64.
4 / cosα = 8.
α = arccos (4/8) = arccos 0.5 = 60º.
Answer: The throw angle is 60º.