Question

1. a ship maneuvers to within 2500 m of an island’s 1800m high mountain peak and fires a projectile at an enemy ship 610 m on the other side of the peak. if the ship shoots the projectile with an initial velocity of 250 m/s at an angle of 75°, (a) what is the range of the projectile? (b) how close to the ship did the projectile land? (c) what is the vertical height of the projectile as it goes over the mountain peak? (d) how close does the projectile come to the mountain peak?

vi=2.50×10² m/s
75.0°
1.80×10³ m
2.50×10³ m
6.10×10² m

Explanation:

Step1: Recall Projectile Motion Formulas

The horizontal and vertical motions of a projectile are independent. The horizontal velocity is \( v_{ix} = v_i \cos\theta \), vertical velocity is \( v_{iy} = v_i \sin\theta \). The time of flight \( T \) (for part a) can be found from vertical motion: \( y = v_{iy}t - \frac{1}{2}gt^2 \), with \( y = 0 \) (landing at same height). The range \( R = v_{ix}T \). For vertical height at a horizontal distance \( x \), find time \( t = \frac{x}{v_{ix}} \), then \( y = v_{iy}t - \frac{1}{2}gt^2 \).

Step2: Solve Part (a) - Range of Projectile

First, calculate \( v_{ix} = 250 \cos75^\circ \), \( v_{iy} = 250 \sin75^\circ \).
\( v_{ix} \approx 250 \times 0.2588 = 64.7 \, \text{m/s} \)
\( v_{iy} \approx 250 \times 0.9659 = 241.475 \, \text{m/s} \)

From vertical motion ( \( y = 0 \) at landing):
\( 0 = v_{iy}T - \frac{1}{2}gT^2 \)
\( T = \frac{2v_{iy}}{g} = \frac{2 \times 241.475}{9.8} \approx 49.28 \, \text{s} \)

Range \( R = v_{ix}T = 64.7 \times 49.28 \approx 3190 \, \text{m} \) (or more accurately, using \( g = 9.81 \)):
\( T = \frac{2 \times 250 \sin75^\circ}{9.81} \approx \frac{482.96}{9.81} \approx 49.23 \, \text{s} \)
\( R = 250 \cos75^\circ \times 49.23 \approx 64.70 \times 49.23 \approx 3185 \, \text{m} \approx 3.19 \times 10^3 \, \text{m} \)

Step3: Solve Part (b) - Distance from Ship to Landing

The total horizontal distance from the ship is the range \( R \). From part (a), \( R \approx 3185 \, \text{m} \). Check if the enemy ship is at \( 2500 + 610 = 3110 \, \text{m} \). Wait, maybe miscalculation? Wait, no—wait, the projectile lands at the enemy ship's side? Wait, no, the problem says "fires a projectile at an enemy ship 610 m on the other side of the peak". Wait, maybe I misread. Wait, the ship is 2500 m from the peak, enemy is 610 m beyond the peak. So total distance from firing ship to enemy is \( 2500 + 610 = 3110 \, \text{m} \). But our range calculation gave ~3185 m, which is more. Wait, maybe the landing is at the enemy ship? Wait, no, the vertical motion: maybe the enemy ship is at the same height (sea level), so the range should be the distance to where it lands, which is the enemy ship? Wait, perhaps my initial approach: let's recalculate \( T \) with \( g = 9.81 \):

\( v_{iy} = 250 \sin75^\circ \approx 250 \times 0.9659258 = 241.48145 \, \text{m/s} \)
\( T = \frac{2v_{iy}}{g} = \frac{2 \times 241.48145}{9.81} \approx 49.23 \, \text{s} \)
\( v_{ix} = 250 \cos75^\circ \approx 250 \times 0.2588190 = 64.70475 \, \text{m/s} \)
\( R = 64.70475 \times 49.23 \approx 3185 \, \text{m} \). The distance from the ship to the enemy is \( 2500 + 610 = 3110 \, \text{m} \). Wait, maybe the projectile lands beyond? No, perhaps the problem is that the enemy ship is at the same elevation, so the range is the distance to where it lands, which is calculated as above. But maybe I made a mistake. Alternatively, maybe the vertical displacement at \( x = 3110 \, \text{m} \) is zero? Let's check time at \( x = 3110 \): \( t = \frac{3110}{64.70475} \approx 48.06 \, \text{s} \). Then vertical position: \( y = 241.48145 \times 48.06 - 0.5 \times 9.81 \times (48.06)^2 \). Calculate:
\( 241.48145 \times 48.06 \approx 11606 \)
\( 0.5 \times 9.81 \times 48.06^2 \approx 4.905 \times 2309.76 \approx 11320 \)
\( y \approx 11606 - 11320 = 286 \, \text{m} \). Not zero. So the projectile lands beyond the enemy ship? Wait, the problem says "fires a projectile at an enemy ship", maybe the enemy ship is at the same height, so our range calculation is correct, and the distance from the ship is the range, ~3190 m. But maybe the question (b) is how close to the ship did it land, i.e., the range, or the distance to the enemy? Wait, the problem says: "How close to the ship did the projectile land?" So the landing point's distance from the ship is the range, ~3190 m. But let's check again.

Step4: Solve Part (c) - Vertical Height Over Peak

The horizontal distance to the peak is \( x = 2500 \, \text{m} \). Time to reach peak: \( t = \frac{2500}{v_{ix}} = \frac{2500}{64.70475} \approx 38.64 \, \text{s} \). Vertical height: \( y = v_{iy}t - \frac{1}{2}gt^2 \)
\( y = 241.48145 \times 38.64 - 0.5 \times 9.81 \times (38.64)^2 \)
\( 241.48145 \times 38.64 \approx 9330 \)
\( 0.5 \times 9.81 \times 38.64^2 \approx 4.905 \times 1493 \approx 7320 \)
\( y \approx 9330 - 7320 = 2010 \, \text{m} \). Wait, but the peak is 1800 m high. So the height over the peak is \( 2010 - 1800 = 210 \, \text{m} \)? Wait, no: the vertical height of the projectile as it goes over the peak is \( y \) (its height above sea level), so we need to calculate \( y \) at \( x = 2500 \, \text{m} \). Let's recalculate with more precision:

\( v_{ix} = 250 \cos75^\circ = 250 \times 0.2588190451 = 64.70476128 \, \text{m/s} \)
\( t = 2500 / 64.70476128 ≈ 38.636 \, \text{s} \)
\( v_{iy} = 250 \sin75^\circ = 250 \times 0.9659258263 = 241.4814566 \, \text{m/s} \)
\( y = v_{iy}t - 0.5gt^2 = 241.4814566 \times 38.636 - 0.5 \times 9.81 \times (38.636)^2 \)
Calculate \( 241.4814566 \times 38.636 ≈ 241.4814566 \times 38 + 241.4814566 \times 0.636 ≈ 9176.295 + 153.58 ≈ 9329.875 \)
\( 0.5 \times 9.81 \times (38.636)^2 = 4.905 \times 1492.84 ≈ 4.905 \times 1492.84 ≈ 7322.38 \)
\( y ≈ 9329.875 - 7322.38 ≈ 2007.495 \, \text{m} \). So the projectile's height above sea level at the peak's horizontal position is ~2007 m. The peak is 1800 m, so the height over the peak is \( 2007 - 1800 = 207 \, \text{m} \approx 210 \, \text{m} \).

Step5: Solve Part (d) - Closest Distance to Peak

The closest distance to the peak (which is 1800 m high, at \( x = 2500 \, \text{m} \)) is the vertical distance between the projectile's path and the peak. Wait, no: the peak is a point (2500 m horizontally, 1800 m vertically). The projectile's path is a parabola. The closest distance would be the minimum distance between the projectile's position (at any time) and the peak (2500, 1800). But since the projectile is moving horizontally and vertically, we can consider the vertical distance when it's at \( x = 2500 \, \text{m} \) (since horizontally, that's the peak's position). Wait, no—actually, the peak is at (2500, 1800). The projectile's position at time \( t \) is \( (x(t), y(t)) = (v_{ix}t, v_{iy}t - 0.5gt^2) \). We need to find the minimum distance between \( (v_{ix}t, v_{iy}t - 0.5gt^2) \) and \( (2500, 1800) \). This is a calculus problem: minimize \( d(t) = \sqrt{(v_{ix}t - 2500)^2 + (v_{iy}t - 0.5gt^2 - 1800)^2} \). To simplify, minimize \( d^2(t) \).

Let \( f(t) = (v_{ix}t - 2500)^2 + (v_{iy}t - 0.5gt^2 - 1800)^2 \)
Take derivative \( f’(t) = 2(v_{ix}t - 2500)v_{ix} + 2(v_{iy}t - 0.5gt^2 - 1800)(v_{iy} - gt) \)
Set to zero:
\( (v_{ix}t - 2500)v_{ix} + (v_{iy}t - 0.5gt^2 - 1800)(v_{iy} - gt) = 0 \)

This is a quadratic equation in \( t \). Alternatively, since the peak is at the horizontal distance of 2500 m, maybe the closest approach is at \( t \) when the projectile is at \( x = 2500 \, \text{m} \) (since horizontally, that's the peak's position, and vertically, the projectile is moving up then down; at \( x = 2500 \), the time is \( t = 2500 / v_{ix} ≈ 38.636 \, \text{s} \), as in part (c). At that time, the projectile's height is \( y ≈ 2007 \, \text{m} \), so the vertical distance is \( 2007 - 1800 = 207 \, \text{m} \). But wait, maybe the closest distance is not at the horizontal position of the peak? Let's check the time when the projectile is directly above the peak (horizontally), which is \( t = 2500 / v_{ix} \), as that's when \( x = 2500 \). At that time, the vertical distance is \( y(t) - 1800 \). Since the projectile is moving vertically, before \( t \), it's moving up, after, moving down. At \( t \), it's at the highest point? No, the time to reach maximum height is \( t_{max} = v_{iy}/g ≈ 241.48 / 9.81 ≈ 24.62 \, \text{s} \), which is less than 38.64 s. So at \( t = 38.64 \, \text{s} \), the projectile is on its way down. So the maximum height is at \( t_{max} ≈ 24.62 \, \text{s} \), where \( x_{max} = v_{ix} \times 24.62 ≈ 64.70 \times 24.62 ≈ 1593 \, \text{m} \), which is less than 2500 m. So the projectile passes the peak's horizontal position (2500 m) after reaching maximum height, on the way down. So at \( x = 2500 \, \text{m} \), the projectile is descending. The vertical distance at that point is \( 2007 - 1800 = 207 \, \text{m} \). But is this the closest distance? Let's check the distance at \( t_{max} \): \( x = 1593 \, \text{m} \), \( y = v_{iy}t_{max} - 0.5gt_{max}^2 = 241.48 \times 24.62 - 0.5 \times 9.81 \times (24.62)^2 \). Calculate \( 241.48 \times 24.62 ≈ 5945 \), \( 0.5 \times 9.81 \times 24.62^2 ≈ 4.905 \times 606.1 ≈ 2974 \), so \( y ≈ 5945 - 2974 = 2971 \, \text{m} \). The horizontal distance from peak is \( 2500 - 1593 = 907 \, \text{m} \), vertical distance is \( 2971 - 1800 = 1171 \, \text{m} \). The distance from peak at \( t_{max} \) is \( \sqrt{907^2 + 1171^2} ≈ \sqrt{822649 + 1371241} ≈ \sqrt{2193890} ≈ 1481 \, \text{m} \), which is more than 207 m. So the closest distance is at \( x = 2500 \, \text{m} \), vertical distance ~207 m. So the closest distance to the peak is ~207 m (or 210 m).

Answers (Summary):

(a) Range: ~\( 3.19 \times 10^3 \, \text{m} \) (or 3190 m)
(b) Distance from ship: ~3190 m (or check if enemy is at 3110 m, but range is ~3190 m)
(c) Vertical height over peak: ~2010 m (height above sea), so over peak: ~210 m
(d) Closest distance to peak: ~210 m

(Note: Detailed calculations may vary slightly with \( g \) value. Using \( g = 9.81 \, \text{m/s}^2 \) for precision.)

Answer:

s (Summary):
(a) Range: ~\( 3.19 \times 10^3 \, \text{m} \) (or 3190 m)
(b) Distance from ship: ~3190 m (or check if enemy is at 3110 m, but range is ~3190 m)
(c) Vertical height over peak: ~2010 m (height above sea), so over peak: ~210 m
(d) Closest distance to peak: ~210 m

(Note: Detailed calculations may vary slightly with \( g \) value. Using \( g = 9.81 \, \text{m/s}^2 \) for precision.)