An arrow, launched from a flat horizontal surface of the ground at an angle to the horizon, fell back

| September 11, 2021 | education

An arrow, launched from a flat horizontal surface of the ground at an angle to the horizon, fell back to the ground 40 meters from the place of the throw. What was the speed of the arrow 2 seconds after the throw, if at that moment it was directed horizontally?

Task data: S (boom flight range) = 40 m; t1 (considered moment in time) = 2 s = t / 2, where t is the total duration of the flight (true, since the speed is directed horizontally (the considered arrow has reached its maximum height)).

Let us transform the formula for calculating the flight range: S = V02 * sin2α / g = V02 * 2 * sinα * cosα / g.

Let’s substitute in the formula the flight time of the arrow and the horizontal speed: t = 2V0 * sinα / g and V0x = V0 * cosα.

The formula will take the form: S = (2 * V0 * sinα / g) * (V0 * cosα) = t * V0x = 2t1 * V0x, whence we express: V0x = S / 2t1.

Calculation: V0x = 40 / (2 * 2) = 10 m / s.

Answer: The boom speed was 10 m / s.



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