The illustration for this problem can be seen at screenshots(dot)firefox(dot)com/atwXACAA9akaoAji/null

The army is testing out a new prototype artillery cannon with an uncommonly high muzzle velocity of 1000 m/sec . The design bugs haven't been fully worked out yet, so the cannon has to wait at least a full minute between shots (reloading, etc.). The cannon can be aimed with an angle between 0 and 90 degrees with respect to the horizontal.

A particularly demanding exercise is being conducted, in which the cannon fires an initial shot into the air, and while the first shell is still in the air, a second shot is fired (at a different angle) so that both shells impact the target (at ground level) simultaneously.

What is the maximum horizontal distance from the cannon to the target, such that this is possible?

Details and Assumptions:

Assume level ground, with no air resistance.
The gravitational acceleration is 10 m/s².
Give your answer in meters, to the nearest whole meter.
For the sake of this problem, ignore the Earth's curvature.

For a time lag of k seconds (k>=60),

y1 = 1000sinθ t - 10t^2
x1 = 1000cosθ t

y2 = 1000sinØ (t-k) - 10t^2
x2 = 1000cosØ (t-k)

Since maximum range for y1 is achieved at θ=45°, k must be less than 70.2 seconds.

see what you can do with that.

maximum time in air is when the ball gets to maximum height, or when θ=90°, so the vertical speed is v=100-10t=0 at t=10, making total flight time 20 seconds.

There is no way to wait at least a minute for the 2nd shot while the 1st is still up in the air.

To find the maximum horizontal distance from the cannon to the target, we need to determine the conditions under which both shells will hit the ground simultaneously.

Let's analyze the problem step by step:

1. The first shell is fired at an angle from the horizontal. We know that the maximum height reached by a projectile fired at an angle is achieved when it's fired at a 45-degree angle. However, this angle does not guarantee that the second shell will hit the target simultaneously.

2. We can use the time of flight formula for a projectile fired at an angle to find the time it takes for the first shell to reach the ground. The time of flight (t) is given by the equation:

t = (2 * v * sinθ) / g

Where:
v = initial velocity (1000 m/s)
θ = angle of projection (unknown)
g = acceleration due to gravity (10 m/s²)

3. Once we find t, we know that the second shell needs to be fired within that time to hit the target simultaneously. Therefore, the second shell will have a time of flight (t') less than or equal to t.

4. To find the maximum horizontal distance, we can use the range formula for projectile motion. The range (R) is given by the equation:

R = v * cosθ * t'

Where:
v = initial velocity (1000 m/s)
θ = angle of projection (unknown)
t' = time of flight for the second shell

5. Now, to find the maximum horizontal distance, we need to optimize the equation for R. We can do this by finding the value of θ that maximizes the range R. We can use calculus to find the maximum value of R by taking the derivative of R with respect to θ, setting it equal to zero, and solving for θ.

At this point, we have the steps to find the maximum horizontal distance. However, we still need the value of θ. Unfortunately, the screenshot you provided is not accessible in text format. Therefore, without the specific values for the first shell angle, we cannot provide the exact answer.

To solve the problem completely, you need to know the angle of projection for the first shell. Once you have that information, you can follow the steps outlined above to find the maximum horizontal distance.