In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vigorous 21 m/s, catching the ball at the highest point in his jump. Right after catching the ball, how fast is the receiver moving?

Respuesta :

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

[tex]m_1v_1+m_2v_2 = (m_1+m_2)v_f[/tex]

Where,

[tex]m_{1,2}[/tex]= Mass of each object

[tex]v_{1,2}[/tex] = Initial velocity of each object

[tex]v_f[/tex]= Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

[tex]m_2v_2 = (m_1+m_2)v_f[/tex]

[tex](0.42)(21) = (90+0.42)v_f[/tex]

[tex]v_f = 0.0975m/s[/tex]

Therefore the velocity right after catching the ball is 0.0975m/s

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