The body is thrown at an initial speed of 20 m / s at an angle to the horizontal of 45

The body is thrown at an initial speed of 20 m / s at an angle to the horizontal of 45 degrees. Find the body speed at a height of 7 m above the ground.

To find the speed of the thrown body at the considered height, we use the law of conservation of energy: Ek1 + En1 = Ek0; m * Vx ^ 2/2 + m * g * h1 = m * V0 ^ 2/2, whence we express: Vx = √ (2 * (V0 ^ 2/2 – g * h1)).

Variables and constants: V0 – speed at the moment of throw (V0 = 20 m / s); g – acceleration due to gravity (g ≈ 10 m / s2); h1 – considered height (h1 = 7 m).

Calculation: Vx = √ (2 * (V0 ^ 2/2 – g * h1)) = √ (2 * (20 ^ 2/2 – 10 * 7)) ≈ 16.12 m / s.

Answer: An abandoned body at a height of 7 m should have had a speed of 16.12 m / s.



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