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LUCKY_DIMON [66]
3 years ago
8

S ship maneuvers to within 2.50x10^3 m of an islands 1.80x10^3 m high mountain peak and fires a projectile at an enemy ship 6.10

x10^2 m on the other side of the peak. if the ship shoots projectile with an initial velocity of 2.50x10^2m/s at an angle of 75 degrees, how close vertically does the projectile come to the peak
Physics
1 answer:
lilavasa [31]3 years ago
5 0

velocity of the projectile is given as

v = 2.50 \times 10^3 m/s

now the angle is given as 75 degree

so here we will have

v_x = 2.50 \times 10^3 cos75 = 0.65 \times 10^3 m/s

v_y = 2.50 \times 10^3 sin75 = 2.4 \times 10^3 m/s

now the time taken to reach the peak is given as

t = \frac{x}{v_x}

t = \frac{2.50 \times 10^3}{0.65 \times 10^3} = 3.86 s

now the height moved by the projectile in same time is given as

y = v_y t + \frac{1}{2}at^2

now we have

y = (2.4 \times 10^3)(3.86) - \frac{1}{2}(9.81)(3.86)^2

y = 9.2 \times 10^3 m

so distance between the peak and projectile is given as

d = 9.2 \times 10^3 - 1.80 \times 10^3 = 7.4 \times 10^3 m

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