Question

A particle takes $n$ seconds less and acquires a velocity $um/sec$. higher at one place than at another place in falling through the same distance. Calculate the product of the acceleration due to gravity at these two places.

Answer 1

$t_1=T,t_2=T-n$

$V_1=V,V_2=V+u$

$s=\frac{1}{2}g{t}^2$

$s=\frac{1}{2}{g}_1{T}^2$

$s=\frac{1}{2}{g}_1(T-n{)}^2$

$g_2(T-n{)}^2=g_1{T}^2$

$V^2=2g_{1}s$

$(V+u{)}^2=2g_{2}s$

$\frac{g_1}{g_2}=\frac{V^2}{(V+u{)}^2}$

$\frac{(T-n{)}^2}{T^2}=\frac{V^2}{(V+u{)}^2}$

$\frac{T-n}{T}=\frac{V}{V+u}$

$(V+u)T-(V+u)n=VT$

$T(V+u-V)=(V+u)n$

$uT=(V+u)n$

$T=\frac{(V+u)n}{u}$

$g_2(Tn{)}^2=g_1{T}^2$

$g_2(\frac{(V+u)n}{u}-n{)}^2$=$g_1(\frac{(V+u)n}{u}{)}^2$

$g_2\left(\right(\frac{V}{u})n{)}^2$=$g_1(\frac{(V+u)n}{u}{)}^2$

$g_2\frac{V^2}{u^2}=g_1\frac{(V+u{)}^2}{u^2}$

$\frac{g_1}{g_2}=\frac{V^2}{(V+u{)}^2}$

$V=gt$

$V=g_{1}T$

$V+u=g_2(T-n)$

$T=\frac{(V+u)n}{u}$

$V=g_1\frac{(V+u)n}{u}$

$uV=g_1(Vn+un)$

$=V{g}_{1}n+u{g}_{1}n$

$V(u-g_{1}n)=u{g}_{1}n$

$V=\frac{u{g}_{1}n}{u-g_{1}n}$

$g_1{g}_2=\frac{V}{T}\times \frac{V+u}{T.n}$

$=\frac{V}{\frac{(V+u)n}{u}}\times \frac{V+u}{\frac{V}{u}n}$

$g_1{g}_2=\frac{u^2}{n^2}$