Respuesta :
Answer:
The time it will take the bomb to reach the target is approximately 2.402 seconds
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, [tex]u_v[/tex], of the bomb = 160·√2 m/s downwards
From the equation of motion, we have;
s = [tex]u_v[/tex] × 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;
[tex]x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}[/tex]
[tex]t = \dfrac{-160 \times \sqrt{2} \pm \sqrt{(160 \times \sqrt{2} )^{2}-4\times 9.81\times (-600)}}{2\times 9.81}[/tex]
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.
