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The Zacchini family was renowned for their human- cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of 69 m. He was propelled inside the barrel for 5.2 m and launched at an angle of 53°. If his mass was 85 kg and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at 53°. Neglect air drag.)

Respuesta :

Answer:

ecceleration   a = 67,626 m / s²

Explanation:

We have a projectile launch, where they give us the horizontal reach of man with this we can calculate its initial velocity

       R = Vo² sin 2θ / g

       Vo² = gR / sin 2θ

       Vo = √ (9.8 69 / sin 2 53) = √ 703.45

       Vo = 26.52 m / s

This is the speed with which it leaves the canyon, as they indicate that the acceleration in the canyon is constant, let's calculate the average acceleration

        Vfc² = Voc² + 2 a X

Let's look at the values ​​of the quantities in this part of the problem, Vfc is the speed at the tip of the canyon, that is, it is 26.52 m / s, X is the distance traveled that corresponds to the length of the cannon 5.2 m and Vo is the speed initial before the shot that is worth zero (Vo = 0)

   

         Vfc² = 2 a X

         a = Vfc² / 2X

         a = 26.52²/2 5.2

         a = 67,626 m / s²

Let's use Newton's second law to look for strength

        F = ma

        F = 85 67,626

        F = 5748.2 N

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