velocity height dispacement

Question

1.1. a submarine M sails directly west and lines a torpedo N in the same direction. Immediately after the torpedo N, the submarine M sails W29’S. the submarine M sails at 85 km/h and the speed of the torpedo is 194 km/h. Calculate the velocity of the torpedo N relative to the velocity of the submarine M in magnitude and direction. 1.2. a stone is thrown at a velocity of 74 m/s at an angle of 38 degrees to the horizontal. Determine the following: 1.2.1. the maximum height that the stone reaches 1.2.2. the horizontal displacement of the stone. 1.3. an aeroplane flies to the destination 600km west of its starting point. A wind is blowing from the north-west at 75 m/s. the pilot wants to complete the flight in 55 minutes. Calculate the following: 1.3.1. the direction to be taken by the pilot 1.3.2. the velocity of the aircraft with respect to the ground.

Answer 1

1.2.1)

Given data :

Angle θ = 38°

Initial velocity v₀ =  74 m/s

Earth gravity g = 9.8 m/s

The maximum height reached by the stone H_max = ?

Formula :

Maximum height H_max = (v₀²sin²θ)/2g

Substitute : v₀ =  74 m/s , θ = 38° and  g = 9.8 m/s.

H_max = (74²sin²38) / (2×9.8)

H_max = (5476×0.379) / (19.6)

H_max =105.89 m

Maximum Height  H_max = 105.89 m.

Answer 2

1.2.2)

Given data :

Angle θ = 38°

initial velocity v₀ =  74 m/s

Earth gravity g = 9.8 m/s

The horizontal displacement reached by the stone d = ?

Formula :

Horizontal displacementd = (v₀²sin2θ)/g

Substitute : v₀ =  74 m/s , θ = 38° and  g = 9.8 m/s.

d = (74²sin(2×38)) / (9.8)

d = (5476)(0.97) / (9.8)

d = 542.02 m

The horizontal displacement is 542.02 meters

Answer 3

1.3.2)

Destination distance = 600 km in west direction.

Destination time t_d= 55 minutes

t_d= 55/60 = 0.917 hr

Wind  blowing speed v_w=   75 m/s in north-west direction.

v_w=   75 * 3.6 = 270 km/hr

∠COA = Relative angle

Speed of flight = distance / time = 600 / 0.917 = 654 km/hr

OA = 645 km/hr

OB = 270 km/hr

Resultant speed DE² = OA² + OB² - 2 OA OB cos45

DE² = 645² + 270² - 2 (270)(645) cos45

DE² = 645² + 270² - 2 (270)(645) cos45

DE = 492.58 km/hr

For parallelogram

(diagonal1)² + (diagonal2)² = (side1 + side2)²

(DE)² + (OC)² = (OE + OD)²

(OC)² = ( 645 + 270 )² - 492.58²

OC = 771.5 km/hr.

Solution :

The velocity of the aircraft with respect to the ground is 771.5 km/hr

 

Answer 4

1.3.1)

Destination distance = 600 km in west direction.

Destination time t_d= 55 minutes

t_d= 55/60 = 0.917 hr

Wind  blowing speed v_w=   75 m/s in north-west direction.

v_w=   75 * 3.6 = 270 km/hr

∠COA = Relative angle

Speed of flight = distance / time = 600 / 0.917 = 654 km/hr

OA = 645 km/hr

OB = 270 km/hr

Resultant speed DE² = OA² + OB² - 2 OA OB cos45

DE² = 645² + 270² - 2 (270)(645) cos45

DE² = 645² + 270² - 2 (270)(645) cos45

DE = 492.58 km/hr

For parallelogram

(diagonal1)² + (diagonal2)² = (side1 + side2)²

(DE)² + (OC)² = (OE + OD)²

(OC)² = ( 645 + 270 )² - 492.58²

OC = 771.5 km/hr.

From OAC right angled triangle

cosθ = OA/OC

cosθ = 645/771.5

cosθ = 0.836

θ = 33.3^o

Solution :

Direction of the velocity of plane relative to the velocity of ground is 33.3° upside inclined to west