Distance and Displacement

Question

A ship maneuvers 2500 m of islands 1800 m high mountain peak and fires a projectile at an enemy ship 610 m on the other side of the peak. The initial velocity of the projectile is 2500 and it is launched at an angle of 75. How close to the enemy ship does the projectile land? How close vertically does the projectile come to the peak?

Answer 1

The statement of the problem does not indicate dimensions of the velocity.

I will assume that the dimensions are m/s, although a velocity of 2500 m/s

too unrealistically high, and as a result you will see that it will overshoot by a large amount.

However, I hope that explaining the approach will help you solve the problem with the correct velocity.

Let

s = 2500 m be the distance of the ship from the mountain,

h = 1800 m be the height of the mountain,

d = 610 m be the distance of the enemy ship from the mountain,

v = 2500 m/s be the initial velocity of the projectile,

α = 75° be the angle of launch against the horizontal,

g = 9.81 m/s² be gravitational acceleration,

x(t) be the horizontal distance of the projectile after time t,

y(t) be the vertical distance of the projectile after time t.

We are given

x(0) = 0,

y(0) = 0

Horizontally, the projectile travels at a constant velocity vcos(α), so

x(t) = vcos(α)t (1)

Vertically, the projective travels at a constant velocity vsin(α) while at the same

time it is falling down with acceleration g. Therefore

y(t) = vsin(α)t - gt²/2. (2)

First let's answer the second question because if the projectile does clear

the mountain then the first question is mute.

We can calculate the time t₁ the projectile flies over the maintain, because at that time t1 it satisfies

x(t₁) = s

Substituting (1)

vcos(α)t₁ = s

t₁ = s/vcos(α)

At that time t₁ the height of the projectile is

y(t₁) = vsin(α)s/vcos(α) - g(s/vcos(α))²/2 = s tan(α) - g(s/vcos(α))²/2

Substituting actual values

y(t₁) = 2500×tan(75°) - 9.81×(2500/(2500×cos(75°)))²/2 = 9256.9 m

That is more than 8646 m over the mountain.

The projectile will hit water level at a time t₂ satisfying

y(t₂) = 0

Substituting (2)

vsin(α)t₂ - gt₂²/2 = 0

This equation has two solutions.

The solution t₂ = 0 merely tells us that originally at time 0 the projectile was at water level.

The other solution

t₂ = 2vsin(α)/g

is the time the projectile hits water level at the other side of the mountain.

At that time t₂ the distance from the launch site will be

x(t₂) = vcos(α)t₂ = 2v²sin(α)cos(α)/g = v²sin(2α)/g

Substituting actual numbers

x(t₂) = 2500²×sin(150°)/9.81 = 318552 m

That is more that 315 km away from the target.