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Batman (mass = 88.8 kg) jumps straight down from a bridge into a boat (mass = 530 kg) in which a criminal is fleeing. The velocity of the boat is initially +14.5 m/s. What is the velocity of the boat after Batman lands in it?

Respuesta :

Answer:

The velocity of the boat after Batman lands in it is 2.08 m/s.

Explanation:

Given that,

Mass of batman [tex]m_{1}= 88.8\ kg[/tex]

Mass of boat [tex]m_{2}=530\ kg[/tex]

Initial velocity = 14.5 m/s

We need to calculate the velocity of boat

Using conservation of momentum

[tex]m_{1}u=(m_{1}+m_{2})v[/tex]...(I)

Where, [tex]m_{1}[/tex]=mass of batman

[tex]m_{2}[/tex] =mass of boat

u=initial velocity

v = velocity of boat

Put the value in the equation

[tex]88.8\times14.5=530+88.8\times v[/tex]

[tex]v=\dfrac{88.8\times14.5}{530+88.8}[/tex]

[tex]v=2.08\ m/s[/tex]

Hence, The velocity of the boat after Batman lands in it is 2.08 m/s.

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