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# A Stone is dropped from the top of the tower and travel 24.5 m in last second of its journey. the height of the tower is ...?

8 years ago

44.1m is the height of the tower. assume the time taken to reach the ground 2 b (2s/g)^(1/2) (s is the height of the tower) and then put this in the equation Sn= u+(g/2)*(2n-1) which is the distance travelled the the nth second. here u=0 and take g=9.8

8 years ago

Dear Shubham,
Distance travelled at nth sec=u + a/2(2t-1)
so here 24.5=0+4.9(2t-1)
hence t=3 secs
it takes 3 secs to reach the bottom so height=1/2gt^2
H=4.9x9=44.1 m(ANS)
Regards.

3 years ago
The answers given above with the solution. I`m eager just to know in the answer which says dear shubham.. how did the person use that formula of nth term. That`s it... The derivation from which it came.
2 years ago
Distance travelled in the last second(nth second)=24.5m
Assuming u=0 and g=10 m/s^2
According to the formula,
S(nth)= u +a/2 (2n-1)
24.5 = 0+10/2 (2n-1)
24.5 = 5(2n-1)
24.5 = 10n -5
10n = 29.5
n=2.95
n~3 sec
That means, t=3 s
According to the second kinematical equation,
s = ut + 1/2 at^2
S = o +1/2 ×10× 9
S= 5 ×9
S = 45m
Hence, answer to this question is 45m
(Note: this is an approximate answer. Hence you can choose the option closest to this in the examinations. Remember, approximation is important to save time and solve sums with mininmum errors and hence save time!)
Hope this helped:) Yash Chourasiya
11 months ago
Hello Student

Let the total height of the tower ishand the total time is taken to reach the ground is t.
The height is given as,
h = 1/2​gt2 …........ (1)
The height before last seconds is given as,
h−24.5 = ​0.5 g(t−1)2
0.5gt2 − 24.5 = 0.5​g(t)2 + 0.5​g − gt
t = 3s
Substitute the value oftin the equation (1), we get
h = 0.5×9.8​×(3)2
h = 44.1m
Thus, the height of the tower is 44.1m.