Complete Question 
In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
Answer:
The speed of the helicopter is 
Explanation:
From the question we are told that 
    The height at which he let go of the brief case is  h =  130 m  
     The  time taken before the the brief case hits the water is  t =  6 s 
Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as 
        
 
Here s  is the distance covered by the bag at sea level which is zero
        
 
=>     
 
=>   
=>   
       
 
        
             
        
        
        
Answer:Comparison of the embryological development of different species also reveals similarities that show relationships not evident in the fully-formed anatomy.
Explanation:
 
        
                    
             
        
        
        
To find the ratio of planetary speeds Va/Vb we need the orbital velocity formula: 
V=√({G*M}/R), where G is the gravitational constant, M is the mass of the distant star and R is the distance of the planet from the star it is orbiting. 
So Va/Vb=[√( {G*M}/Ra) ] / [√( {G*M}/Rb) ], in our case Ra = 7.8*Rb 
Va/Vb=[ √( {G*M}/{7.8*Rb} ) ]  / [√( {G*M}/Rb )], we put everything under one square root by the rule: (√a) / (√b) = √(a/b) 
Va/Vb=√ [ { (G*M)/(7.8*Rb) } / { (G*M)/(Rb) } ], when we cancel out G, M and Rb we get:
Va/Vb=√(1/7.8)/(1/1)=√(1/7.8)=0.358 so the ratio of Va/Vb = 0.358.  
        
             
        
        
        
Answer:
D. magnitude and direction
Explanation: