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Grace [21]
3 years ago
5

A ship maneuvers to within 2500 m of an island's 1800 m high mountain peak and fires a projectile at an enemy ship 610 m on the

other side of the peak, as illustrated in Figure 3-29. If the ship shoots the projectile with an initial velocity of v = 248 m/s at an angle of θ = 74°, how close to the enemy ship does the projectile land?
Physics
2 answers:
Ne4ueva [31]3 years ago
3 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

t=(0-(250sin75)^2)/-9.8 
<span>the distance one is (2500+610)- (250m/s*cos75)*t=Dh Dh=horizontal distance </span>

<span>the max height one is d=0.5*9.8*t^2 </span>
<span>d= max height subtract 1800-d</span>
valkas [14]3 years ago
3 0

Answer: 2714.24 m

Explanation:

The projectile would have to cross the mountain peak to reach the other side and hit the enemy ship.

Vertical component of velocity = Vy = 248 sin 74° m/s = 238.4 m/s

horizontal component of velocity = Vx = 248 cos 74° m/s = 68.4 m/s

The time taken to reach the peak by the projectile would be half the time taken by projectile to cover the horizontal distance across the mountain.

Using second equation of motion:

Δy = Vy t + 0.5 a t²

⇒0 = 238.4 m/s × t + 0.5 × (-9.8 m/s²) t²

t = 238.4/4.9 = 48.6 s

Horizontal distance covered, x = Vx (t) = 68.4 m/s × 48.6 s = 3324.24 m

It lies at (3324.24 - 610) m = 2714.24 m close to the ship.

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