A particle is dropped from a tower it is found that it travel 45 metres in the last second of journey find out the height of the tower
Answers
Answered by
3
Distance travelled in last second
S = u + a(n - 1/2)
45 = 0 + 10(n - 1/2)
n = 4.5 + 0.5
n = 5th second
Total time of flight is 5 seconds
Height of the tower
H = 0.5gt^2
= 0.5 × 10 × 5^2
= 125 m
Height of the tower is 125 m
S = u + a(n - 1/2)
45 = 0 + 10(n - 1/2)
n = 4.5 + 0.5
n = 5th second
Total time of flight is 5 seconds
Height of the tower
H = 0.5gt^2
= 0.5 × 10 × 5^2
= 125 m
Height of the tower is 125 m
Prajwal35:
Thanks friend
Welcome
BTW why did you reported my answer?
Similar questions