Complete Question
The complete question is shown on the first uploaded image
Answer:
The velocity of the bus at B is [tex]v_B= 31.6\ m/s[/tex]
Explanation:
Let's take position B as base point .
From the diagram height between point B and A ia mathematically evaluated as
[tex]h_A = 60 -55[/tex]
[tex]h_A = 5 \ m[/tex]
From the question we are told that
The velocity at location A is [tex]v_A = 30 m/s[/tex]
According the law of conservation of energy
[tex]PE_A + KE_A = KE_B[/tex]
Where [tex]PE_A[/tex] is the potential energy at A which is mathematically represented as
[tex]PE_A = mgh_A[/tex]
[tex]KE_A[/tex] is the kinetic energy energy at A which is mathematically represented as
[tex]KE_A = \frac{1}{2} mv^2_A[/tex]
[tex]KE_B[/tex] is the kinetic energy at A which is mathematically represented as
[tex]KE_B = \frac{1}{2} mv_B^2[/tex]
Where [tex]v_B[/tex] is the velocity at location B
So
[tex]mgh_A + \frac{1}{2} mv_A^2 = \frac{1}{2} mv_B^2[/tex]
Making [tex]v_B[/tex] the subject of the formula
[tex]v_B= \sqrt{\frac{gh_A + \frac{1}{2} v_A^2 }{0.5} }[/tex]
Substituting values
[tex]v_B= \sqrt{\frac{(9.8 * 5) + \frac{1}{2} 30^2 }{0.5} }[/tex]
[tex]v_B= 31.6\ m/s[/tex]
