A ship maneuvers to within 2500 m of an island's 1800 m high mountain peak and fires a projectile at an enemy ship 610 m on the other side of the peak, as illustrated in Figure 3-29. If the ship shoots the projectile with an initial velocity of v = 248 m/s at an angle of θ = 74°, how close to the enemy ship does the projectile land?
Well, it seems like the ship was trying to play a game of "hide and seek" with the enemy ship, but ended up playing a game of "shoot and miss" instead.
Now, let's calculate how close the projectile lands to the enemy ship.
First, we need to break down the initial velocity into its horizontal and vertical components. The horizontal component is given by v * cos(θ), and the vertical component is given by v * sin(θ).
The time it takes for the projectile to reach the peak of the mountain can be calculated using the vertical component of velocity and the acceleration due to gravity. Trust me, gravity always seems to interfere with everyone's plans.
Anyway, once we know the time it takes to reach the peak, we can calculate the horizontal distance traveled during that time using the horizontal component of velocity.
Now, the projectile travels a distance of 1800 m vertically up the mountain peak. So, subtracting this distance from the distance traveled horizontally gives us the remaining horizontal distance to the enemy ship.
Finally, we can add up this remaining horizontal distance to the 610 m on the other side of the peak to find out how close the projectile lands to the enemy ship.
And voila! We should have our answer. But don't worry, even if the ship missed the target, at least they had a good view of the mountain peak during their failed attempt.
To find out how close the projectile lands to the enemy ship, we can break down the problem into horizontal and vertical components.
First, we need to find out how long it takes for the projectile to reach the peak of the mountain. We can use the vertical motion equation:
y = y0 + v0y * t - 0.5 * g * t^2
Where:
y = vertical displacement (1800 m)
y0 = initial vertical position (0 m)
v0y = vertical component of the initial velocity (v * sin(θ))
g = acceleration due to gravity (-9.8 m/s²)
t = time
Since the initial vertical position is 0, the equation simplifies to:
y = v0y * t - 0.5 * g * t^2
Rearranging the equation, we get:
0.5 * g * t^2 - v0y * t + y = 0
This equation is a quadratic equation in terms of time (t), so we can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where:
a = 0.5 * g
b = -v0y
c = y
Next, we can find the time it takes for the projectile to travel from the peak of the mountain to the enemy ship by using the horizontal motion equation:
x = x0 + v0x * t
Where:
x = horizontal displacement (610 m)
x0 = initial horizontal position (0 m)
v0x = horizontal component of the initial velocity (v * cos(θ))
t = time
Since the initial horizontal position is 0, the equation simplifies to:
x = v0x * t
Now, we can calculate the horizontal component of the initial velocity (v0x) and the vertical component of the initial velocity (v0y):
v0x = v * cos(θ)
v0y = v * sin(θ)
Now, substitute the values into the equations:
v0x = 248 m/s * cos(74°)
v0y = 248 m/s * sin(74°)
Now that we have the values for v0x and v0y, we can substitute them into the equations above to find the time it takes for the projectile to reach the peak of the mountain (t1) and the time it takes to travel from the peak to the enemy ship (t2).
Finally, calculate the horizontal displacement of the projectile when it lands close to the enemy ship:
x = x0 + v0x * t2
Substitute the values into the equation and calculate x to find out how close the projectile lands to the enemy ship.
find time in air:
hf=hi+viv*t-4.9t^2
0=0+248sin74 t-4.9t^2
t=248sin74/4.9
now, with that time in air.
horizontal range=248cos74*t
figure that out, then since the distance to the ship is 610+2500
you can figure the distance the projectile is from the ship.