question_answer

A particle is thrown with velocity u making an angle q with the vertical. It just crosses the top of two poles each of height h after 1 s and 3 s respectively. Find the maximum height of projectile \[\text{g}=\text{9}.\text{8m}/{{\text{s}}^{\text{2}}}\].

Answer:

                Let the projectile go from O to P in time \[{{\text{t}}_{\text{1}}}\left( =\text{1s} \right)\]and from O to Q in time \[{{\text{t}}_{\text{2}}}=\left( \text{3s} \right)\] Initial vertical velocity of particle at \[\text{O}=\text{ucos}\theta \]. Taking vertical upward motion of particle from O to P and O to Q, we have   \[\text{h}=\text{ucos}\theta {{\text{t}}_{\text{1}}}-\frac{1}{2}gt_{1}^{2}=\text{ucos}\theta {{\text{t}}_{\text{2}}}\frac{1}{2}gt_{2}^{2}\] or            \[\text{u cos}\theta \times \text{1}\frac{1}{2}\text{g}\times {{\text{1}}^{\text{2}}}=\text{ucos}\theta \times \text{3}~\frac{1}{2}\text{g}{{\text{3}}^{\text{2}}}\] or            \[\text{u cos}\theta (\text{3}-\text{1})=~\frac{g}{2}\left( \text{9}\text{1} \right)=~\frac{9.8}{2}\times \text{8}=\text{4}.\text{9}\times \text{8}\] or            \[\text{u cos}\theta =\text{ }\frac{\text{4}.\text{9}\times \text{8}}{2}\times \text{4}=\text{19}.\text{6 m}/\text{s}\] Max. height, \[\text{H}=\frac{{{u}^{2}}{{\cos }^{2}}\theta }{2g}=\frac{{{(19.6)}^{2}}}{2\times 9.6}=19.6m\]