Question

A body projected obliquely with velocity $19.6m/s$ has its kinetic energy at the maximum height equal to $3$ times its potential energy. Since projection from the ground, its height from ground after $1s$ is
($h=$ maximum height, take $g=9.8m/s^2)$.

Answer 1

Initial velocity , $u=19.6m/s$

after every 1 sec of projectile angle made by particle with horizontal is given by

$t=\frac{usin\theta }{g}$

$1=\frac{19.6\times sin\theta }{9.8}$

$sin\theta =\frac{9.8}{19.6}$

$sin\theta =\frac{1}{2}$

$\theta ={30}^0$

Therefore position of particle after 1 sec is given by

$H_{max}=\frac{u^{2}si{n}^2\theta }{2g}$

$=\frac{19.6\times 19.6\times si{n}^2{30}^0}{2\times 9.8}$

$=19.6\times \frac{1}{4}$

$=4.9m$