Question

An aircraft flying horizontally with a velocity of 540 km/h at a height of 2000 m releases a bomb from it.Find (a) the time taken by the bomb to reach the ground, (b) the velocity with which it hits the target on the ground and (c) distance of tthe target

Answer 1

Lets convert the velocity to m/s

V = 540 × 5/18 = 150 m/s

we will use the following kinematics equation.

S = ut + 0.5gt²

Take g = 10

Doing the substitution:

2000 = 150t + 5t²

Divide through by 5:

400 = 30t + t²

t² + 30t -400 = 0

Solving the quadratic equation:

The roots are :

-10 and + 40

Doing the substitution we have :

t² - 10t + 40t - 400 = 0

t(t - 10) + 40 ( t - 10) = 0

(t + 40) (t - 10) = 0

Hence :

t = -40 or 10

We take 10 since time is positive:

The time is thus 10 seconds.

The final velocity is given by:

v = u + gt

V = 150 + 10 × 10 = 250 m/s

= 250 m/s

The distance of the target is given by:

Distance of the target is Range.

Range = ut

This equals:

150 × 10 = 1500 m