a particle is dropped from the top of a tower.it covers 40m in last 2 seconds.find the height of the tower
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40 =u+10/2
u =35
time of descent = 3.5
s = 35×3.5 + 10×( 3.5×3.5) /2
u =35
time of descent = 3.5
s = 35×3.5 + 10×( 3.5×3.5) /2
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The distance travelled by the particle in nth sec
Snth = u + a ( 2n - 1) / 2
Here, Snth = 40 m , u = 0 , a = 10 m/s2
so, 40 = 10 ( 2n - 1) / 2
or, n = 4.5 sec
So, the particle takes 4.5 s to reach the ground from top of the tower.
Now, the height of the tower
h = ut + gt2 /2
= 0 + 10 (4.5 )2 /2
= 101.25 m