Question

A particle having a mass $05kg$ is projected under gravity with a speed of $98m/s$ at an angle of ${60}^o$ . The magnitude of the change in momentum in $Kgm/s$ of the particle after $10s$ . Find range, height and time of flight ?

Answer 1

initial velocity is $\overrightarrow{u}=98\cos \theta i+98\sin \theta j$

Finally-$\overrightarrow{v}=98\cos \theta i+(98\sin \theta -gt)j$

$\Delta \overrightarrow{p}=m(\overrightarrow{v}-\overrightarrow{u})=m(-gt)j$

$⟹\Delta p=5\times 9.8\times 10=490kgm/s$

$R=\frac{u^2\sin 2\theta }{g}=\frac{{98}^2\sqrt{3}}{2\times 9.8}=848.7m$

$H=\frac{u^2{\sin }^2\theta }{2g}=\frac{{98}^2\times 3}{4\times 2\times 9.8}=367.5m$

$T=\frac{2u\sin \theta }{g}=\frac{2\times 98\times \sqrt{3}}{2\times 9.8}=17.32s$