question_answer

An object is thrown vertically upward with some speed. It crosses 2 points p, q which are separated by h metre. If t is the time between p and highest point and coming back and \[{{\text{t}}_{\text{q}}}\] is the time between q and highest point and coming back, relate acceleration due to gravity, \[{{\text{t}}_{p}}\], \[{{\text{t}}_{\text{q}}}\]and h.

Answer:

                Acceleration, \[\text{a}\,\text{=}\frac{dv}{dt}=-\text{r}{{\omega }^{\text{2}}}\text{sin}\omega \text{t}\]or d\[dv=-\text{r}{{\omega }^{\text{2}}}\text{sin}\omega \text{t}\,dt\]; Integrating it we have  \[\text{v}=-\text{r}\omega \left( -\frac{\cos \omega t}{\omega } \right)=\text{ r}\omega \,\,\,\text{cos}\omega \text{t}\]. Now, velocity \[\text{v}=\frac{dx}{dt}=\text{ r}\omega \text{ }\,\text{cos}\omega \text{t}\]or \[\text{dx }=\text{ r}\omega \,\text{cos}\omega \text{t dt}\] Integrating it, we have, \[\text{x}=\text{r}\omega \,\left( \frac{\sin \omega t}{\omega } \right)=\text{r sin}\omega \text{t}\].