A small drone launched at an angle of $45^\circ$ with horizontal moves with constant velocity of $\sqrt {200}$ m/s. Its power shuts down just 4 second after the launch. Find the horizontal range of the drone. [Ignore air, $g=10 m/s^2$]
Solution
Motion with constant velocity
$d = ucos\theta.t = 40m$
$h = usin\theta.t = 40m$
Projectile motion
$y = -40 = usin\theta .t - \frac {1}{2} g t^2 = 10.t-\frac {1}{2}.10.t^2=10t-5t^2$
$\therefore 5t^2-10t-40=0$
$\Rightarrow t^2-2t-8=0$
$\Rightarrow (t-4)(t+2) = 0$, t = 4 sec
$R=ucos\theta .t=10.4 = 40 m$
Range = d+R = 40+40 m = 80 m