Question

Particle is projected with speed $5\sqrt{2}\text{m/s}$ at an angle of projection ${45}^{\circ }$ from horizontal. At maximum height it breaks in two identical particle, velocity of one particle is zero, find time (in sec) when first part strikes the ground after explosion. Take $g=10{\text{m/s}}^2$

Answer 1

$H=\frac{(5\sqrt{2}{)}^2{\sin }^2\theta }{2g}=\frac{50\times 1}{20\times 2}=\frac{50}{40}=\frac{5}{4}m$
$T=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times \frac{5}{4}}{10}}=\frac{1}{\sqrt{4}}=0.5\text{sec}$