A ball is dropped from height of 45 metre then distance covered by it in last second of its motion will be
Answers
Explanation:
s = ut + 1/2 at^2 (here u is the initial velocity of the ball)
Since the initial velocity was 0 m/s, the formula becomes s = 1/2 at^2.
(substituting the values in the formula)
45 = 1/2 x 9.8 x t^2
45 = 9.8/2 x t^2
45 = 4.9 x t^2 (9.8/2 = 4.9)
(divide 4.9 on both sides)
45/4.9 = 4.9 x t^2 / 4.9
9.1836 = t^2
3.030 = t
Answer:
Explanation:
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.