Answer:
Let h be the height of a tower and u be the initial
velocity of projection of the ball.
(i) When ball is projected vertically upwards. Taking
vertically downwards motion of ball from top of tower to ground we have
\[\text{u}=-\text{u};\,\,\text{a}=\text{g};\,\,\text{s}=\text{h};\text{t}=\text{
}{{\text{t}}_{\text{1}}}\]
As \[\text{s}=\text{ut}+\frac{1}{2}~\text{a}{{\text{t}}^{\text{2}}}\]
\[\therefore \] \[\text{h
}=-\text{u}{{\text{t}}_{\text{1}}}+~\frac{1}{2}\times \text{g }\times
t_{1}^{2}\] ...(i)
(ii) When ball is projected vertically downwards, Taking
vertically downward motion of ball from top of tower to ground, we have \[\text{u
}=\text{ u},\text{a }=\text{ g };\text{ s }=\text{ h};\text{t
}={{\text{t}}_{\text{2}}}\]
\[\therefore \] \[\text{h }=\text{
u}{{\text{t}}_{\text{2}}}+\frac{1}{2}\text{gt}_{2}^{2}\] ...(ii)
(iii) When ball falls freely, u = 0 ; a = g ;
s = h; t = ?; \[\text{h}=~\frac{1}{2}\text{g}{{\text{t}}^{\text{2}}}\] ...(iii)
Multiplying (i) by \[{{\text{t}}_{\text{2}}}\]and (ii) by \[{{\text{t}}_{\text{1}}}\]and
then adding, we get
\[\text{h}\left( {{\text{t}}_{\text{1}}}+\text{
}{{\text{t}}_{\text{2}}} \right)\text{ }=\frac{1}{2}\text{g
}{{\text{t}}_{\text{1}}}{{\text{t}}_{\text{2}}}\left( {{\text{t}}_{\text{1}}}+\text{
}{{\text{t}}_{\text{2}}} \right)\] or \[\text{h
}=\frac{1}{2}\text{g }{{\text{t}}_{\text{1}}}{{\text{t}}_{\text{2}}}\]
Putting this value in (iii), we get
\[\frac{1}{2}\text{g}{{\text{t}}_{\text{1}}}{{\text{t}}_{\text{2}}}=\text{g}{{\text{t}}^{\text{2}}}\]or
\[{{\text{t}}^{\text{2}}}=\text{
}{{\text{t}}_{\text{1}}}{{\text{t}}_{\text{2}}}\]or \[\text{t}=\sqrt{{{t}_{1}}{{t}_{2}}}\]
Putting this value of t in (i), and simplifying it, we get
\[{{\text{L}}_{\text{min}}}=\frac{{{d}_{1}}{{v}_{2}}-{{d}_{2}}{{v}_{1}}}{\sqrt{v_{1}^{2}+v_{2}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
