Physics, 30.10.2019 05:31 dondre54

Aship 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

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Answer from: avaverretkj

202.81 m

490.88 m

Explanation:

$v_{o}$ = 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

=vertical displacement = 0 m

= 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

= horizontal displacement

= 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

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

= closest distance from the ship

d = x - D

d = 3311.81 - 3109

d = 202.81 m

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

= horizontal displacement = 2500 m

= 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

= vertical displacement = ?

= 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

= height of the mountain = 1780 m

= the closest distance of projectile from peak

h = y - H

h = 2270.88 - 1780

h = 490.88 m

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Aship maneuvers to within 2.50 x 103 m of an island's 1.78 x 103 m high mountain peak and fires a pr...

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