A small drone is launched from a helicopter flying at a height of 720 m and moving horizontally at a constant velocity of 180 km/h. The velocity of the drone is 90 km/h with respect to the helicopter having same direction as that of helicopter and remains constant until it loses power after flying for 4 minutes. How far away from its launching point will the drone land? [Ignore air resistance. Take $g=10 m/s^2$.]
Solution
Velocity of drone w.r.t. the ground when it is launched=180+90=270 kmph
Horizontal distance covered in 4 minutes = 270 kmph . 4 min. = 18 km
Time to fall t after it loses power can be obtained from the equation,
$720 m = \frac {1}{2} gt^2$
$144 = t^2$
t = 12 sec
Horizontal distance covered after it loses power = 270 kmph. 12 sec = 0.9 km
Total horizontal distance = 18 km + 0.9 km = 18.9 km