For this assignment, you submit answers by questions. You are required to use a

Question

For this assignment, you submit answers by questions. You are required to use a new randomization after every 3 question submissions Assignment scoring Your best submission for each question part is used for your score. My Notes O Ask Your Teacher 1. -12 points SerPSET9 4 P 089 06 submissions Used An enemy ship is on the east side of a mountain island, as shown in the figure. The enemy ship has maneuvered to within di 2095 m of the h 1490 m high mountain peak and can shoot projectiles with an initial speed of vi 249 m/s. If the western shoreline is horizontally d2 266 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship? m, or more than less than Need Help? Read It Submit Answer Save progress

Explanation / Answer

Mountain height, h = 1490 m
Distance between enemy ship and mountain, d1 = 2095 m
Distance between western shoreline and mountain, d2 = 266 m
Initial speed of projectile, vi = 249 m/s

We need H and L to solve this problem. From the figure H will set the highest elevation that will clear the mountain and given the closest range, while L will set the lowest elevation that will clear the mountain and given the farthest range. To find these angles, we need to use displacement equations in the horizontal and vertical directions.

Horizontal displacemen:
x = vi*cosq*t
t = x/(vi*cosq)

Vertical displacement:
y = vi*sinq*t - (1/2)gt^2
y = vi*sinq*t - (1/2)gt^2
y = vi*sinq*[x/(vi*cosq)] - (1/2)g[x/(vi*cosq)]^2
y = xtanq - [(gx^2/2vi^2)*(1/cos^2^q)]

Since sin^2^q + cos^2^q = 1, then divide is by cos^2^q, we get
tan^2^q + 1 = 1/cos^2^q
This is the quadratic equation in tanq.
Solve this equation, we get
qH = 75.6 degrees and qL = 50.1 degrees

Range at qH:

R = vi^2*sin2qH/g
R = (249)^2*sin(2*75.6)/9.8
R = 3047.88 m
Closest distance to the shore is determined by

3047.88 m - 266 m - 2095 m = 686.88 m

Range at qL:

R = vi^2*sin2qL/g
R = (249)^2*sin(2*50.1)/9.8
R = 6226.64 m

6226.64 m - 266 m - 2095 m = 3865.64 m

So, the safe distances are less than 686.88 m or greater than 3865.64 m from the shore.

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