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A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 28 m/s when it reaches ground level. part a what was its speed when its height was half that of its starting point?

Respuesta :

Let the distance be s 
initial velocity was u = 0m/s  
final velocity was v = 28m/s     
Since no acceleration is provided, only acceleration here is due to gravity. So here, a = 9.81m/s2    

Using Newtons motion equation, 

v^2 = u^2 + 2as     
28^2 = 0 + 2 x 9.81 x s          
s = 784/19.62          
s = 81.55m           
s/2 = 39.96m   
         
Now,        
same equation, to find v at s/2      
v^2 = 0 + 2 x 9.81 x 39.96      
v^2 = 784.0152     
v = 28m/s            

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