To solve this problem, we can break down the shell's motion into its horizontal and vertical components.
Given: Initial speed (u) = 45 m/s Launch angle (θ) = 30° Height of the hill (h) = 400 m Acceleration due to gravity (g) = 9.8 m/s²
First, we need to find the time of flight (t) of the shell. The time it takes for the shell to reach the ground can be determined using the vertical component of its motion. We'll use the following formula:
h = ut * sin(θ) - (1/2) * g * t²
Substituting the known values:
400 = 45 * t * sin(30°) - (1/2) * 9.8 * t²
Simplifying the equation:
400 = 22.5 * t - 4.9 * t²
Rearranging the equation:
4.9 * t² - 22.5 * t + 400 = 0
Solving this quadratic equation, we find two possible solutions for t: t ≈ 8.17 s and t ≈ 12.83 s. Since we're interested in the time of flight, we choose the positive solution: t ≈ 12.83 s.
Now, we can find the horizontal distance traveled by the shell (range). We'll use the formula:
range = u * cos(θ) * t
Substituting the known values:
range = 45 * cos(30°) * 12.83
Simplifying the equation:
range ≈ 45 * 0.866 * 12.83 ≈ 532.57 m
Therefore, the shell will land approximately 532.57 meters from the base of the hill.
To find the angle at which the shell will land, we can use the tangent of the launch angle. Let's call it φ.
tan(φ) = (height of the hill) / (range)
tan(φ) = 400 / 532.57
Taking the inverse tangent (arctan) of both sides:
φ ≈ arctan(400 / 532.57)
Evaluating the expression:
φ ≈ 37.4°
Therefore, the shell will land approximately 532.57 meters away from the base of the hill and at an angle of approximately 37.4° from the horizontal.