A missile is fired at an initial velocity of 1500m/s. What should be the angle to reach a maximum horizontal target? (Show proof)
I need your help.
the max range is reached at an angle of 45°.
Review the equation for range:
v^2/g sin(2θ)
now it is clear that 45° gives max range.
To calculate the angle at which a missile should be fired to reach a maximum horizontal target, we need to consider the projectile motion equations and principles.
1. In projectile motion, the horizontal and vertical components of motion are independent of each other. So, we can analyze them separately.
2. The horizontal component of motion remains constant throughout and can be represented by the formula:
Horizontal Distance (d) = Initial Velocity (v₀) * Time of Flight (t)
3. The vertical component of motion can be analyzed using the equation of motion for vertical displacement:
Vertical Displacement (h) = (Initial Vertical Velocity (v₀) * Time (t)) - (0.5 * Acceleration due to Gravity (g) * Time (t)²)
4. The time of flight can be calculated using the equation:
Time of Flight (t) = (2 * Initial Vertical Velocity (v₀) * sin(θ)) / g
Here, θ is the launch angle and g is the acceleration due to gravity (approximately 9.8 m/s²).
5. To reach the maximum horizontal distance, the time of flight should be maximized.
So, we differentiate the equation for time of flight with respect to θ and equate it to zero:
d(t)/d(θ) = (2 * Initial Vertical Velocity (v₀) * cos(θ)) / g = 0
6. Solving this equation, we find that cos(θ) = 0, which implies θ = 90 degrees or π/2 radians.
However, this angle corresponds to vertical launch, and we need to maximize horizontal distance. So, we choose the complementary angle, which is 0 degrees or 0 radians.
Therefore, the angle required to reach the maximum horizontal target is 0 degrees or 0 radians.
In summary, to maximize the horizontal distance, the missile should be fired at an angle of 0 degrees or 0 radians (horizontal launch).