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Theivanpabloescorcia
11.10.2019 •
Physics
Aship maneuvers to within 2.50 x 10^3 m of an islands 1.80 x 10^3 m high mountain peak and fires a projectile at an enemy ship 6.10 x 10^2 m on the other side of the peak. if the ship shoots the projectile with an inital velocity of 2.50 x 10^2 m/s at an angle of 75.0 degrees, how close to the enemy ship does the projectile land? how close (vertically) does the projectile come to the peak?
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Ответ:
Explanation:
Distance between ship and enemy ship
= 500 + 610
= 3110 m
Range of projectile
R = u² sin2θ / g
= (250x 250 sin 150) / 9.8
= 3188m
The projectile falls within a distance of 3188 - 3110 = 78 m from enemy ship
Height of mountain = 1800 m
We shall find the height of projectile when its horizontal displacement is 2500m
x = 2500 , y = ?
u = 2500 ,
y = x / cos θ - .5 g x² /u²cos² θ
9660 - 7315 m
= 2345 m
It is within 545 m from mountain peak .
Ответ: