37. A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is (g = 10 m/s2)

37. A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is (\(g = 10 m/s^2\))

(a) 60 m

(b) 45 m

(c) 80 m

(d) 50 m

Answer :  (b) 45 m

\( s = ut + \)\(\frac{1}{2}\)\(at^2 \)

\( 40 = u_2(2) + \)\(\frac{1}{2}\)\((10)(2)^2 \)

\( 40 = 2u_2 + 20 \)

\(\rightarrow u_2 = 10 \, m/s \)

\( v^2 = u^2 + 2as \)

\( 10^2 = 0 + 2(10)h \)

\( h = 5 \, m \)

total height of the tower = 40 + 5 = 45 m