velocity of the projectile is given as
[tex]v = 2.50 \times 10^3 m/s[/tex]
now the angle is given as 75 degree
so here we will have
[tex]v_x = 2.50 \times 10^3 cos75 = 0.65 \times 10^3 m/s[/tex]
[tex]v_y = 2.50 \times 10^3 sin75 = 2.4 \times 10^3 m/s[/tex]
now the time taken to reach the peak is given as
[tex]t = \frac{x}{v_x}[/tex]
[tex]t = \frac{2.50 \times 10^3}{0.65 \times 10^3} = 3.86 s[/tex]
now the height moved by the projectile in same time is given as
[tex]y = v_y t + \frac{1}{2}at^2[/tex]
now we have
[tex]y = (2.4 \times 10^3)(3.86) - \frac{1}{2}(9.81)(3.86)^2[/tex]
[tex]y = 9.2 \times 10^3 m[/tex]
so distance between the peak and projectile is given as
[tex]d = 9.2 \times 10^3 - 1.80 \times 10^3 = 7.4 \times 10^3 m[/tex]
The projectile traveled past the height of the enemy's ship.
The given parameters;
The maximum height traveled by the projectile is calculated as follows;
[tex]H = \frac{u^2 sin^2 \theta}{2g} \\\\H = \frac{(250)^2 \times (sin \ 75)^2}{2\times 9.8} \\\\H = 2,975.63 \ m[/tex]
Total vertical height traveled by the projectile is 2,975.63 m.
The position of the enemy's ship on the other side of the peak = 610 m
Thus, we can conclude that the projectile traveled past the height of the enemy's ship.
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