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A soccer ball is kicked with a speed of 35 m/s at an angle of 30° with respect to the ground and lands a certain distance away from the point it was kicked. If a second ball were kicked from the same place with a speed of 37 m/s, at what angle with respect to the ground could its initial velocity be such that it lands in the same place as the first ball?

Respuesta :

Answer:

θ = 25.4º

Explanation:

For this exercise we can use the projectile launch equations, let's use the scope ratio

      R = vo² sin 2θ / g

For the first ball

     R1 = 35² 2 sin (2 30) /9.8

     R1 = 108.25 m

For the second ball,  they ask that the second ball fall in the same place, so the horizontal distance is the same

    R2 = R1

Therefore we calculate the angle

    sin 2θ = R1 g / vo²

    sin 2θ = 108.25 9.8 / 37²

    sin 2θ = 0.7749

    2θ = sin⁻¹ (0.7749)

    2θ = 50.8º

    θ = 25.4º

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