Question

A ship maneuvers to within 2.50 x 103 m of an island's 1.78 x 103 m high mountain peak and fires a projectile at an enemy ship 6.09 x 102 m on the other side of the peak. The ship shoots the projectile with an initial velocity of 2.54 x 102 m/s at an angle of 74.9 degress. The acceleration of gravity is 9.81 m/s2.

a) How close to the enemy ship does the projectile land? Answer in units of m

b) How close (vertically) does the projectile come to the peak? Answer in unity of m

Answer 1

Answer:

a)

$202.81$ m

b)

$490.88$ m

Explanation:

a)

$v_{o}$ = initial velocity of launch = 254 ms⁻¹

$\theta$ = Angle of launch = 74.9 deg

Consider the motion of the projectile along the vertical direction

$v_{oy}$ = initial velocity of projectile along vertical direction = $v_{o} Sin\theta$  = $254 Sin74.9$ = 245.23 ms⁻¹

$a_{y}$ = acceleration = - 9.8 ms⁻²

$y$ =vertical displacement = 0 m

$t$ = time of travel = ?

Using the kinematics equation

$y = v_{oy} t + (0.5)a_{y} t^{2}$

inserting the values

$0 = (245.23) t + (0.5) (- 9.8) t^{2}$

$t = 50.05$ s

Consider the motion of the projectile along the horizontal direction

$v_{ox}$ = initial velocity of projectile along horizontal direction = $v_{o} Cos\theta$  = $254 Cos74.9$ = 66.17 ms⁻¹

$a_{o}$ = acceleration = 0 ms⁻²

$x$ = horizontal displacement

$t$ = time of travel = 50.05 s

Using the kinematics equation

$x = v_{ox} t + (0.5)a_{x} t^{2}$

$x = (66.17) (50.05) + (0.5) (0) (50.05)^{2}$

$x = 3311.81$ m

$D$ = distance of the enemy ship from the point of launch = 2500 + 609 = 3109 m

$d$ = closest distance from the ship

$d = x - D$

$d = 3311.81 - 3109$

$d = 202.81$ m

b)

Consider the motion of the projectile along the horizontal direction from the point of launch to till it reach the peak of mountain

$v_{ox}$ = initial velocity of projectile along horizontal direction = $v_{o} Cos\theta$  = $254 Cos74.9$ = 66.17 ms⁻¹

$a_{o}$ = acceleration = 0 ms⁻²

$x$ = horizontal displacement = 2500 m

$t$ = time of travel = ?

Using the kinematics equation

$x = v_{ox} t + (0.5)a_{x} t^{2}$

$2500 = (66.17) t + (0.5) (0) t^{2}$

$t = 37.78$ s

Consider the motion of the projectile along the vertical direction from the point of launch to till it reach the peak of mountain

$v_{oy}$ = initial velocity of projectile along vertical direction = $v_{o} Sin\theta$  = $254 Sin74.9$ = 245.23 ms⁻¹

$a_{y}$ = acceleration = - 9.8 ms⁻²

$y$ = vertical displacement = ?

$t$ = time of travel = 37.78 s

Using the kinematics equation

$y = v_{oy} t + (0.5)a_{y} t^{2}$

$y = (245.23) (37.78) + (0.5) (- 9.8) (37.78)^{2}$

$y = 2270.88$ m

$H$ = height of the mountain = 1780 m

$h$ = the closest distance of projectile from peak

$h = y - H$

$h = 2270.88 - 1780$

$h = 490.88$ m

Answer 2

We have that from the Question, it can be said that The distance b/w the projectile and the enemy ship and How close the projectile comes to the peak

• d=-199m  
• dp=1285m

From the Question we are told

A ship maneuvers to within 2.50 x 103 m of an island's 1.78 x 103 m high mountain peak and fires a projectile at an enemy ship 6.09 x 102 m on the other side of the peak. The ship shoots the projectile with an initial velocity of 2.54 x 102 m/s at an angle of 74.9 degrees. The acceleration of gravity is 9.81 m/s2.

a) How close to the enemy ship does the projectile land? Answer in units of m

b) How close (vertically) does the projectile come to the peak?

a)

Generally how close in distance it is in ship two is

$d=2.5*10^3+6.09*10^2-(3.308*10^3)\\\\d=-199m$

Therefore

The distance b/w the projectile and the enemy ship is

d=-199m

b)

Generally the equation for the Height is mathematically given as

$H=\frac{u^2sin^2\theta}{2g}\\\\H=(2.54*10^2)*sin^274\\\\H=$

Therefore

The peak distance is

$dp=3065-1780$

dp=1285m

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