Question

A bomber is diving towards it's target at an angle of 45 degrees below the horizontal and at a speed of 320 m/s. When the bomber is 600m above the ground, it releases its load, which then hits the target. How long will it take the bomb to reach the target?

Answer 1

The time it will take the bomb to reach the target is approximately 2.402 seconds

Step-by-step explanation:

The given information are;

The direction the bomber is travelling = 45° below the horizontal

The speed at which the bomber was travelling = 320 m/s

The elevation at which the bomber releases its load = 600 m

Therefore, we have;

The height at which the bomb is released, s = 600 m

The inclination of the direction of motion of the bomb = 45° below the horizontal

The magnitude of the velocity of the bomb at release = 320 m/s

The vertical component of the velocity = 320 m/s × sin(45°) = 160·√2 m/s

Therefore, the initial vertical velocity,
`u_v`, of the bomb = 160·√2 m/s downwards

From the equation of motion, we have;

s =
`u_v` × t + 1/2 × g × t²

Where;

t = The time it takes the bomb to hit the target

g = The acceleration due to gravity = 9.81 m/s²

∴ 600 = 160·√2 × t + 9.81 × t²

9.81 × t² + 160·√2 × t - 600 = 0

By the quadratic formula, we have;

`x = \frac{-b\pm \sqrt{b^(2)-4\cdot a\cdot c}}{2\cdot a}`

`t = \frac{-160 * √(2) \pm \sqrt{(160 * √(2) )^(2)-4* 9.81* (-600)}}{2* 9.81}`

t ≈ -25.467 seconds or t ≈ 2.402 seconds

Given that t is a natural number, we have the correct option is t ≈ 2.402 seconds

Therefore, the time it will take the bomb to reach the target ≈ 2.402 seconds.