Question

A helicopter flying horizontally at a height of 1.960 kilometres moves at a velocity of 600 kilometres per hour. At the time when the aircraft is vertically at point X on earth's surface, a bomb is dropped from it, which hits point Y on the ground. Calculate the distance XY covered.

Answer 1

given ,
H = 1.96 km
v = 600 km/h

this question based on concept of horizontal projectile . so ,bomb moves in the path of projectile .

so , velocity in x -axis always constant
and variable in y -axis

use ,
x = Uxt + 1/2at²
= Uxt + 0
= 600× t

now,
but time taken to strike the ground is
t = √(2H/g)
so,

distance covered in x-axis ( horizontal ) = Ux√(2H/g)
= 600 × √(2×1.96/{ 9.8×10^-3×(3600)² }

=600 ×1/3600×√(1960×2/9.8)
=1/6 × 20
= 20/6 = 10/3 = 3.33 km