Prove that the flight range of two bodies thrown with equal modulus speeds is the same

| September 10, 2021 | education

Prove that the flight range of two bodies thrown with equal modulus speeds is the same if the sum of the angles at which they are thrown to the horizon is 90 °.

The flight range of a body L thrown at a speed V at an angle to the horizon alpha is calculated by the formula L = V ^ 2 * sin (2 * alpha) / g, where g is the acceleration of gravity.

According to the condition of the problem, the flight range of the first body is:

L1 = V ^ 2 * sin (2 * alpha) / g.

Flight range of the second body:

L2 = V ^ 2 * sin (2 * (90 – alpha)) / g = V ^ 2 * sin (180 – 2 * alpha) / g.

Let us simplify the expression sin (180 – 2 * alpha) in the second equation by the rule of the sum of two angles.

sin (180 – 2 * alpha) = sin (180) * cos (2 * alpha) – cos (180) * sin (2 * alpha) = 0 * cos (2 * alpha) – (-1) * sin (2 * alpha) = sin (2 * alpha).

Hence L1 = L2.



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