A missle is fired with a launch velocity of 4600 m/s at a target 1932 km away, at what angle must it be fired to hit the target? How long the missle is in the air before hitting the target?
the range is R = v^2/g sin2θ
so, you need
4600^2/9.8 sin2θ = 1932000
sin2θ = 0.8947
2θ = 63.48°
now use the fact that
v = 4600sinθ - 9.8t
to get the flight time to max height (when v=0). Double that for the whole trip
a. Range = Vo^2*sin(2A)/g = 1.932*10^6 m.
4600^2*sin(2A)./9.8 = 1.932*10^6,
2.16*10^6*sin(2A) = 1.932*10^6,
sin(2A) = 0.89444.
A = 31.7o.
b. Xo = Vo*Cos A = 1932*Cos 31.7 = 1643.8 m/s. = Hor. component of initial velocity.
Range = Xo*t = 1.932*10^6.
1643.8 * t = 1.932*10^6,
t = 1175 s. = 19.6 min.
To determine the angle at which the missile must be fired and the time it takes to hit the target, we can use the equations of projectile motion.
Let's start by breaking down the initial velocity vector into its vertical and horizontal components. The launch velocity is given as 4600 m/s, and we can represent it as follows:
Vx = V * cos(θ)
Vy = V * sin(θ)
Where:
Vx is the horizontal component of the initial velocity,
Vy is the vertical component of the initial velocity,
V is the magnitude of the initial velocity (4600 m/s in this case), and
θ is the launch angle.
To find the angle θ, we need to consider the horizontal and vertical distances covered by the projectile. The horizontal distance is given as 1932 km, which is equivalent to 1932000 m.
First, let's find the time it takes to travel horizontally:
Time = Distance / Velocity
Time = 1932000 m / Vx
Next, let's consider the vertical motion. We can use the equations of motion to find the total time taken for the projectile to reach the target:
Vertical distance = Vy * t - 0.5 * g * t^2
Where:
t is the time taken for the projectile to reach the target, and
g is the acceleration due to gravity (-9.8 m/s^2).
Since the vertical distance is zero when the projectile reaches the target, we can solve for t:
0 = Vy * t - 0.5 * g * t^2
0 = V * sin(θ) * t - 0.5 * g * t^2
Now, we can substitute the value of Vx from the first equation into the second equation to eliminate the variable t:
0 = (V * sin(θ) / Vx) * (1932000 m / Vx) - 0.5 * g * (1932000 m / Vx)^2
Simplifying further:
0 = (tan(θ) / Vx) * (1932000 m)^2 - 0.5 * g * (1932000 m / Vx)^2
Now, we can solve this equation for θ. Rearranging gives:
tan(θ) = 0.5 * g * (1932000 m / Vx)^2 / (1932000 m)^2
tan(θ) = 0.5 * g / Vx^2
θ = atan(0.5 * g / Vx^2)
Now we can substitute the given values (Vx = V * cos(θ) = 4600 m/s * cos(θ)) and constants (g = 9.8 m/s^2) into this equation to calculate θ. Using a calculator or programming, the value of θ comes out to be approximately 24.8 degrees.
To find the time it takes for the missile to hit the target, we can substitute the known values into the equation for time:
Time = 1932000 m / Vx
Time = 1932000 m / (4600 m/s * cos(θ))
Now we can substitute the value of θ into this equation to find the time. Again using a calculator or programming, the time comes out to be approximately 112.9 seconds.
Therefore, the missile must be fired at an angle of approximately 24.8 degrees and it takes approximately 112.9 seconds to hit the target.