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Example 34 A particle is dropped from the top of a tower.Its displacement in the first three seconds and in the lastsecond is the same. Find the height of the tower.

Answer»

Let the height of the tower be H...so, total time is t = √2H/g

Displacement in 1st three seconds is

h3 = 1/2*g*(3)² = 9g/2 =45m

also displacement in nth second is

hn = g/2*(2√(2H/g)-1)

but h3 = hn

so, 45 = 5*(2*(√(2H/g) -1)=> 9 = 2*√H/5 -1 => 10 = 2√H/5=> 5 = √H/5 => H/5 = 25

so, H = 25*5 = 125m



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