A mortar fires a projectile at an angle of of 30 degrees above the horizontal with a velocity of 500 m/s. A soldier spots a tank to be travelling in the direction of the mortar at a constant the rate of 4 m/s. How far should the tank be from the mortar before the mortar is fired in order to hit the tank?

Vo = 500m/s[30o]

Xo = 500*cos30 = 433 m/s.
Yo = 500*sin30 = 250 m/s.

Tr = -Yo/g = -250/-9.8 = 25.5 s. = Rise
time.

Tf = Tr = 25.5 s. = Fall time.

Tr+Tf = 25.5 + 25.5 = 51 s. = Time in air.

Dx = Xo*(Tr+Tf) = 433*51 = 22,083 m. =
Distance travelled by the shell.

D = Dx + Dt = 22,083 + 4m/s*51s. = 22,287 m. = Distance from mortar to tank.

17.35

To solve this problem, we can break it down into horizontal and vertical components.

1. Horizontal Component:
The horizontal velocity of the projectile remains constant throughout the motion since there is no force acting on it horizontally. In this case, the horizontal component of the projectile's velocity is given by:
Vx = V * cos(θ)
Vx = 500 m/s * cos(30°)
Vx = 500 m/s * √(3) / 2
Vx ≈ 433.0 m/s

2. Vertical Component:
The vertical component of the projectile's velocity changes due to the acceleration due to gravity. In this case, the vertical component of the projectile's velocity is given by:
Vy = V * sin(θ)
Vy = 500 m/s * sin(30°)
Vy = 500 m/s * 0.5
Vy = 250 m/s

3. Time of Flight:
The time it takes for the projectile to hit the ground can be determined using the vertical component of the projectile's velocity. Since the mortar is fired at an angle of 30 degrees above the horizontal, we can use the kinematic equation:
Vy = gt
where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time.
250 m/s = 9.8 m/s² * t
t = 250 m/s / 9.8 m/s²
t ≈ 25.51 s

4. Horizontal Distance:
The horizontal distance the projectile travels can be found by multiplying the horizontal component of velocity by the time of flight. In this case:
Distance = Vx * t
Distance = 433.0 m/s * 25.51 s
Distance ≈ 11,054 m

Therefore, the tank should be approximately 11,054 meters away from the mortar before the mortar is fired in order to hit the tank.

To determine the distance between the mortar and the tank before firing, we need to calculate the time it takes for the projectile fired from the mortar to reach the tank.

Let's break down the given information:
Angle of projection (theta) = 30 degrees
Initial velocity (u) = 500 m/s
Horizontal velocity (uh) = ?
Vertical velocity (uv) = ?
Rate of the tank (v) = 4 m/s

Since we are interested in the horizontal distance, we need to find the horizontal velocity (uh). Since the projectile is fired at an angle, we can use trigonometry to relate the horizontal and vertical velocities.

The horizontal velocity (uh) is given by:
uh = u * cos(theta)

Using the given values:
uh = 500 m/s * cos(30 degrees)
uh ≈ 500 m/s * 0.866 (rounded to three decimal places)
uh ≈ 433 m/s (rounded to three decimal places)

Now that we have the horizontal velocity of the projectile, we can determine the time it takes for the projectile to reach the tank. Since the tank is moving in the same direction as the mortar, we subtract the tank's velocity from the horizontal velocity of the projectile.

Relative horizontal velocity (ur) = uh - v
ur = 433 m/s - 4 m/s
ur ≈ 429 m/s

To calculate the time taken to reach the target:
Time (t) = horizontal distance traveled (d) / relative horizontal velocity (ur)

We need to determine the horizontal distance traveled by the projectile. Since the tank is moving continuously during the flight time (t), the horizontal distance traveled by the projectile (d) should equal the distance the tank travels.

Thus, we can rewrite the equation as:
d = v * t

Substituting the value of d into the previous equation:
t = d / ur

Now, we need to consider the vertical motion of the projectile. The vertical distance traveled by the projectile can be calculated using the following equation:

Vertical distance (y) = uv * t + (0.5 * g * t^2)

In this case, since the tank is at the same vertical level as the mortar, the vertical distance (y) will be zero. Therefore, we can set this equation to zero and solve for t:

0 = uv * t + (0.5 * g * t^2)

Since the projectile is fired at an angle, we need to separate the initial velocity (u) into horizontal and vertical components (uh and uv). The vertical velocity can be calculated as:

uv = u * sin(theta)

Using the given values:
uv = 500 m/s * sin(30 degrees)
uv ≈ 500 m/s * 0.5 (rounded to three decimal places)
uv ≈ 250 m/s (rounded to three decimal places)

Considering that the acceleration due to gravity (g) is approximately 9.8 m/s^2, we can substitute these values into the equation and solve for t.

0 = 250 m/s * t - (0.5 * 9.8 m/s^2 * t^2)

Simplifying the equation:
-4.9 t^2 + 250 t = 0

Factoring out 't':
t(-4.9t + 250) = 0

At this point, we have two possible solutions:
1. t = 0 (this corresponds to the initial time when the projectile is fired)
2. -4.9t + 250 = 0

Solving for t, we find:
-4.9t + 250 = 0
4.9t = 250
t = 250 / 4.9
t ≈ 51.02 seconds (rounded to two decimal places)

Since we are interested in the time it takes for the projectile to reach the tank, we discard the initial time of t = 0.

Finally, substituting the value of t into the equation for the horizontal distance:
d = v * t
d = 4 m/s * 51.02 s
d ≈ 204.08 m (rounded to two decimal places)

Therefore, the tank should be approximately 204.08 meters away from the mortar before firing in order to hit the tank.