Question

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 6.10x10^2 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

Answer 1

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

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