Question

what is the distance if a = 2.25m/s^2 v = 45m/s and u = 0m/s

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Answer 1

Answer:When the stone is dropped, initial velocity = 0.

Now, h=(1/2)gt  

2

 

Distance travelled in t  

n

​  

 sec can be written as  h  

n

​  

=(1/2)gt  

n

​  

 

2

 

Distance travelled in t  

n

​  

−1 sec can be written as  h  

n−1

​  

=(1/2)g(t  

n

​  

−1)  

2

 

So, h  

n

​  

−h  

n−1

​  

 = journey in nth second = (1/2)g(t  

n

2

​  

−(t  

n

​  

−1)  

2

)

        =(1/2)g(2t  

n

​  

−1)=g(t  

n

​  

−1/2)

Now using h=(1/2)gt  

2

 for h = 45m we get t = 3 sec

So, in the 3rd second the stone will travel 10 X 2.5 m = 25 m.

Explanation:

Answer 2

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