An enemy plane is flying horizontally at an altitude of 2.0km with a speed of 340m/s.An

army man with anti-craft gun on the ground
sight the enemy with a muzzle speed at 60m/s
.when the enemy plane is directly
overhead.find
a)at what angle from the verticals should the
gun be fired so as to hit the plane.
b)at what minimum altitude should the enemy
fly to avoid being hit(take g=10m/s^2)

Vi = initial up speed = ? cos A

u = constant horizontal component = ? sin A
s = horizontal airplane speed=340
T = flight time for both

horizontal problem:
340 T = u T = ? T sin A

so sin A = 340/(? ) which is kind of obvious :)

now the vertical problem:
h = 0 + Vi t -.49 t^2

To solve this problem, we need to break it down into two parts:

a) Finding the angle at which the gun should be fired to hit the plane
b) Finding the minimum altitude for the enemy plane to avoid being hit

a) Finding the angle:
In order to hit the enemy plane, the projectile launched from the ground needs to travel horizontally and reach the same height as the plane. This means that the vertical velocity component of the projectile must be zero when it reaches the plane's altitude.

Let's assume that the angle from the vertical at which the gun should be fired is θ. The horizontal velocity component of the projectile is the muzzle speed, which is 60m/s in this case.

The vertical velocity component of the projectile can be calculated using the equation:
v_vertical = v_initial * sin(θ)

For the projectile to reach the same altitude as the plane, the time of flight should be the same for both the plane and the projectile. We can calculate the time of flight for the projectile using the equation:
time = distance / velocity

The distance the projectile needs to travel horizontally is the same as the distance between the plane and the ground. Given that the plane is directly overhead and at an altitude of 2.0km, the distance is 2.0km = 2000m.

Therefore, the time of flight for the projectile is:
time = 2000m / 60m/s

Now, we can calculate the vertical displacement of the projectile during that time using the equation of motion:
displacement_vertical = v_vertical * time + (1/2) * g * t^2

Since the vertical component of the projectile's velocity is zero at the plane's altitude, the vertical displacement should be zero.

Setting the displacement to zero and solving for θ, we get:
0 = v_initial * sin(θ) * time + (1/2) * g * time^2

Substituting the given values:
2000m = (60m/s) * sin(θ) * (2000m / 60m/s) + (1/2) * (10m/s^2) * (2000m / 60m/s)^2

Solving this equation will give us the angle at which the gun should be fired to hit the plane.

b) Finding the minimum altitude for the enemy plane:
To calculate the minimum altitude for the enemy plane to avoid being hit, we need to analyze the maximum height reached by the projectile fired from the ground at the given muzzle speed.

The maximum height can be determined using the equation:
maximum height = (initial velocity^2 * sin^2(θ)) / (2 * g)

Substituting the given values:
maximum height = (60m/s)^2 * sin^2(θ) / (2 * 10m/s^2)

Setting the maximum height greater than the altitude of the plane at 2.0km, we can solve for θ to determine the minimum angle at which the enemy plane should fly to avoid being hit.

By following these steps and performing the calculations, you can find the answers to the given problem.