Question

A bomber is flying horizontal with a velocity of 540km/hr. at a hight of 2000m . how far from the energy camp should it releases a bomb so as to hit the camp .(g_10m/s)​

Answer 1

Answer:

Since the bomber drops the bomb from a height of 78.4m, the time taken for the bomb to reach the ground will be given by s=ut+ 21 at 2

where,u is the initial velocity in the vertical direction which is 0

s is the vertical displacement from the ground which is −78.4m

a is the acceleration due to gravity g=−9.8ms

− 2

Upon substituting these values we get t=4s

Since the flight is moving in the horizontal direction with a speed of 150m/s, the distance covered by the bomb in the horizontal direction by the time it reaches the ground is given by (150∗4)m which is equal to 600m

∴The bomb should be released 600m away from the target

Answer 2

Given:-

• Initial Velocity = 0m/s

• Height given = 2000m

• Acceleration due to gravity = + 10m/s

To Find:-

• The Distance from which bomb should be released to hit the camp.

Formulae used:-

• S = ut + ½ × a × t²

Where,

S = Distance

u = Initial Velocity

a = Acceleration

t = Time.

Now,

→ S = ut + ½ × a × t²

→ 2000 = 0 × t + ½ × 10 × t²

→ 2000 = 0+5t²

→ 2000 = 5t²

→ 5t² = 2000

→ t² = 2000/5

→ t² = 400

→ √t² = √400

→ t = 20s.

Since, The flight moving in a horizontal direction with a speed of 150m/s the distance covered by bomb is the distance bomber reaches the ground so.

→ Distance = Speed × time

→ Distance = 150 × 20 = 3000m

Hence, The bomb should ne released from 3000m as to hit the bomber.