a gun can shell targets 15 km away. find

(a)the min launch speed needed to achieve that range
(b) the shells flight time under these conditions

To find the answers, we need to use the equations of projectile motion.

(a) To find the minimum launch speed needed to achieve a range of 15 km, we can use the range formula:

Range = (initial launch speed)^2 * sin(2θ) / g,

where:
- Range is the horizontal distance traveled by the projectile (15 km or 15,000 m in this case),
- θ is the launch angle, and
- g is the acceleration due to gravity (approximately 9.8 m/s^2).

In this case, we assume the launch angle θ is 45 degrees. So, let's plug in the values into the formula and rearrange it to find the initial launch speed:

15,000 = V^2 * sin(90°) / 9.8,
V^2 = (15,000 * 9.8) / sin(90°),
V^2 = 147,000,
V ≈ √147,000,
V ≈ 383 m/s.

Therefore, the minimum launch speed needed to achieve a range of 15 km is approximately 383 m/s.

(b) To find the shell's flight time under these conditions, we can use the time of flight formula:

Time of Flight = 2 * (initial launch speed) * sin(θ) / g.

Using the same launch angle θ of 45 degrees, we can plug in the values:

Time of Flight = 2 * 383 * sin(45°) / 9.8,
Time of Flight ≈ 86 seconds.

Therefore, the shell's flight time under these conditions is approximately 86 seconds.