Question

A particle is projected from ground level with an initial velocity of $35m/s$ at an angle of ${\tan }^{-1}3/4$ to the horizontal. Find the time for which the particle is more than $2m$ above the ground. $[g=10m/s^2]$

Answer 1

The angle is given as,

$\tan \theta =\frac{3}{4}$

$\theta ={36.86}^{\circ }$

The height is given as,

$h=\left(u\sin \theta \right)t-\frac{1}{2}g{t}^2$

$2=21t-5t^2$

$5t^2-21t+2=0$

The time can be written as,

$t_1+t_2=\frac{21}{5}$

$t_1{t}_2=\frac{2}{5}$

${\left(t_1-t_2\right)}^2={\left(t_1+t_2\right)}^2-4t_1{t}_2$

${\left(t_1-t_2\right)}^2={\left(\frac{21}{5}\right)}^2-4\times \frac{2}{5}$

${\left(t_1-t_2\right)}^2={\left(\frac{21}{5}\right)}^2-4\times \frac{2}{5}$

$\left(t_1-t_2\right)=4\sec$

Thus, the time for which the particle is more than $2m$ above the ground is $4\sec$.