Question

answer is 45m. plse give me explanation​

Answer 1

Let the body have velocities $\sf{v_B}$ and $\sf{v_C}$ at positions B and C respectively. The body has zero velocity at A.

Let the body travel 25 metres in last second after travelling $\sf{x}$ metres. So the height of the tower is $\sf{H=x+25.}$

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(0,1){45}}\put(-20,0){\line(1,0){40}}\put(-20,45){\line(1,0){20}}\put(-4.5,22.5){\sf{H}}\put(-3,27){\vector(0,1){18}}\put(-3,21){\vector(0,-1){21}}\put(4,12.5){\sf{25\ m}}\put(8,16){\vector(0,1){9}}\put(8,11){\vector(0,-1){11}}\put(7,35){\sf{x}}\put(8,33){\vector(0,-1){8}}\put(8,38){\vector(0,1){7}}\put(1.5,46){\sf{A}}\put(1.5,26){\sf{B}}\put(1.5,1){\sf{C}}\multiput(0,0)(0,20){2}{\multiput(0,25)(4,0){5}{\line(1,0){2}}}\multiput(-20,0)(2,0){20}{\qbezier(0,-1)(0.5,-0.5)(1,0)}\end{picture}$

The distance 25 metres travelled in last second is given by second equation of motion as,

$\sf{\longrightarrow 25=v_B(1)+\dfrac{1}{2}\,g(1)^2}$

$\sf{\longrightarrow 25=v_B+\dfrac{g}{2}}$

$\sf{\longrightarrow v_B=25-\dfrac{g}{2}}$

Taking $\sf{g=10\ m\,s^{-2},}$

$\sf{\longrightarrow v_B=25-\dfrac{10}{2}}$

$\sf{\longrightarrow v_B=20\ m\,s^{-1}}$

The velocity $\sf{v_B}$ is given by third equation of motion as,

$\sf{\longrightarrow (v_B)^2=0^2+2gx}$

$\sf{\longrightarrow (v_B)^2=2gx}$

$\sf{\longrightarrow 20^2=20x}$

$\sf{\longrightarrow 400=20x}$

$\sf{\longrightarrow x=20\ m}$

Thus height of the tower is,

$\sf{\longrightarrow H=x+25}$

$\sf{\longrightarrow H=20+25}$

$\sf{\longrightarrow\underline{\underline{H=45\ m}}}$

Hence (b) is the answer.