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A body is dropped from some height. In last second it covers 24.5m distance. Calculate the height from which the ball has been dropped.
[ take g = 9.8m/s^2]
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Answers
Hey there !
Solution :
Let the height of the tower be ' h ' and the total time taken be ' t '.
We know that
s = ut + 1 / 2 at²
Here u = 0, hence we can ignore that part. Hence we get,
s = at² / 2
The distance traveled is the height of the tower. Hence s can also be denoted as h. Acceleration involved here is the gravity. Hence the acceleration is 9.8 m / s²
Since the distance of the object at the last second is given we take time to be t seconds out of which in time ( t - 1 ) second it has traveled h - 24.5 m.
So applying in the formula we get,
h - 24.5 = a * ( t - 1 )² / 2
h - 24.5 m = a * ( t² - 2t + 1 ) / 2 -----( Equation 1 )
We know that,
h = at² / 2 -----( Equation 2 )
Substituting this in equation 1 we get,
( at² / 2 ) - 24.5 = a * ( t² - 2t + 1 ) / 2
( at² / 2 ) - 24.5 = ( at² - 2at + a ) / 2
1 / 2 can be written as 0.5. hence we get,
0.5 at² - 24.5 = 0.5 at² - 1 at + 0.5 a
0.5 at² gets cancelled on both sides. Hence we get,
- 24.5 = - at + 0.5 a
at - 24.5 = 0.5 a
We know a = 9.8. Hence applying the values we get,
9.8 t - 24.5 = 4.9
9.8 t = 4.9 + 24.5
9.8 t = 29.4
=> t = 29.4 / 9.8 = 3 seconds
Substitute the time to be 3 in equation 2. We get,
h = at² / 2
h = 9.8 * ( 3 )² / 2
h = 9.8 * 9 / 2
h = 88.2 / 2
h = 44.1 m
Hence the height of the tower is 44.1 m
Hope my answer helped :-)
Total time taken to reach the ground = T
H = 0.5gT^2 ——(1)
H - 24.5 = 0.5g * (T - 1)^2
0.5gT^2 - 24.5 = 0.5g * (T^2 + 1 - 2T)
0.5gT^2 - 24.5 = 0.5gT^2 + 0.5g - gT
gT - 24.5 = 0.5g
10T - 24.5 = 5
10T = 29.5
T = 2.95 s
Substitute T = 2.95 s in equation (1)
H = 0.5gT^2
= 0.5 * 10 * (2.95)^2
= 43.5 m
Height from which the ball has been dropped.is 43.5 m
I hope its help you
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