Answer:
Let u'
be the velocity of the object while crossing point p and v' be its velocity
while crossing point q. Fig. 2(CF).53. A is the highest point of vertical
motion of object. As per question, the time taken by the object in going from p
to \[\text{A}={{\text{t}}_{p}}_{/\text{2}}\] the time taken by the object in
going from q to\[\text{A }={{\text{t}}_{\text{q}}}_{/\text{2}}\].
Taking vertical upward motion of object from p to A, we
have
\[\text{u
}=\text{ u}'\text{,}\upsilon =0,\text{ a}=-\text{g},\text{
t}={{\text{t}}_{\text{p}/\text{2}}}\]
As, \[\upsilon =u+\text{at}\]
\[0\text{ }=\text{ u}'\text{ }+\text{ }\left( -\text{g}
\right){{\text{t}}_{\text{p}/\text{2}}}\]or\[\text{u}'\text{ }=\text{
g}{{\text{t}}_{\text{p}/\text{2}}}\] ...(i)
Taking vertical upward motion of object from q to A, we have,
\[\text{u
}=\text{ }\upsilon ',\text{ }\upsilon \text{ }=\text{ }0,\text{ a }=\text{
}-\text{g},\text{ t }=\text{ }{{\text{t}}_{\text{q}/}}_{\text{2}}\]
As \[\upsilon =u+\text{at}\]
\[0=\upsilon '+\left( -\text{g}
\right){{\text{t}}_{\text{q}/\text{2}}}\]or\[\upsilon '\text{ }=\text{
g}{{\text{t}}_{\text{q}/\text{2}}}\]
Taking vertical upward motion of object from p to q, we have,
\[\text{u}=\text{u}',\text{
v}=\text{v}',\text{ a}=-\text{g},\text{ s}=\text{h}\]
As,\[{{\upsilon }^{\text{2}}}=\text{
}{{\text{u}}^{\text{2}}}+\text{ 2as}\]\[\therefore \]\[\upsilon
{{'}^{\text{2}}}=\text{ u}{{'}^{\text{2}}}+\text{ 2 }\left( -\text{ g}
\right)\text{h}\]
or\[\text{2 gh }=\text{ u}{{'}^{\text{2}}}-\text{ }\upsilon
{{'}^{\text{2}}}=\frac{{{g}^{2}}t_{p}^{2}}{4}-\frac{{{g}^{2}}t_{q}^{2}}{4}\] or\[g=\frac{8h}{(t_{p}^{2}-t_{q}^{2})}\]
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