A ship maneuvers to within 2500m of an island's 1800m high mountain peak and fires a projectile at an enemy ship 610m on the other side of the peak. If the ship shoots the projectile with an initial velocity of 250m/s at an angle of 75 degrees. how close to the enemy ship does the projectile land? How close (vertically)does the projectile come to the peak?
To determine how close the projectile lands to the enemy ship and how close it comes vertically to the mountain peak, we can break down the problem into horizontal and vertical components.
First, let's calculate the time taken for the projectile to reach its highest point:
Time of flight to peak (T₁):
Using the vertical motion equation:
H = V₀₀ * t + 0.5 * g * t²
Where H is the vertical displacement (1800m), V₀₀ is the vertical initial velocity (V₀₀ = V₀ * sin(θ)), g is the acceleration due to gravity (-9.8 m/s²), and t is the time taken to reach the peak.
Plugging in the given values:
1800m = (250 m/s * sin(75°)) * t + 0.5 * (-9.8 m/s²) * t²
Solving this quadratic equation will give us the time T₁ to reach the peak.
Next, let's calculate the time taken for the projectile to reach the horizontal distance of 610m:
Time of flight to target (T₂):
Using the horizontal motion equation:
D = V₀ * cos(θ) * t + 0.5 * a * t²
Where D is the horizontal displacement (610m), V₀ is the initial velocity (250 m/s), θ is the launch angle (75°), and a is the horizontal acceleration (0 since there is no acceleration in the horizontal direction).
Plugging in the given values:
610m = 250 m/s * cos(75°) * t + 0.5 * 0 * t²
610m = 250 m/s * cos(75°) * t
Solving this equation will give us the time T₂ to reach the target.
Finally, to determine how close the projectile lands to the enemy ship horizontally, we calculate the horizontal distance traveled:
Horizontal distance (D):
Using the equation:
D = V₀ * cos(θ) * T₂
Where D is the horizontal distance (our required answer), V₀ is the initial velocity (250 m/s), θ is the launch angle (75°), and T₂ is the time taken to reach the horizontal distance of 610m.
Plugging in the given values:
D = 250 m/s * cos(75°) * T₂
To calculate how close the projectile comes vertically to the mountain peak, we use the vertical motion equation at the time T₁:
Vertical distance from peak (H'):
Using the equation:
H' = V₀₀ * t + 0.5 * g * t²
Where H' is the vertical displacement (our required answer), V₀₀ is the vertical initial velocity (V₀₀ = V₀ * sin(θ)), g is the acceleration due to gravity (-9.8 m/s²), and t is the time taken to reach the peak (T₁).
Plugging in the given values:
H' = (250 m/s * sin(75°)) * T₁ + 0.5 * (-9.8 m/s²) * T₁²
Solving this equation will give us the vertical distance (H') from the peak.
By substituting the values of T₂ and T₁ that we calculated into the equations for D and H', respectively, we can find the horizontal distance and vertical distance from the peak.