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Grade: 11

                        

The distance travelled by the freely falling particle during first 3 seconds is equal to that during last second of its fall from the top of a tower. What is the height of the tower

3 years ago

Answers : (1)

Arun
24742 Points
							
S = u*t + (1/2) a* t²
S = 0+ 5* 9 = 45 m
 
Now
St = u + (1/2) a(2t-1)
 = 0 + 5(2t-1)
 
But 45 = 5(2t-1)
t = 5 sec
Now
H = ut + (1/2 )*a * t²
H = 0 + 5*25
H= 125 metres
3 years ago
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