velocity dispacement height

Question

1.2 A bullet is fired at an angle of 29° to horizontal at a velocity of 410 m/s. Calculate the following: 1.2.1 The maximum height reached by the bullet 1.2.2 The horizontal displacement when the bullet hits the ground 1.3 Two vehicles start moving simultaneously at a fork in a road. Vehicle V travels at a speed of 125 km/h north-east. Vehicle W travels at 125 km/h directly east. Calculate the velocity of vehicle W relative to the velocity of vehicle V in magnitude and direction.

Answer 1

1.2.1)

Given data :

fired angle θ = 29°

initial velocity v₀ =  410 m/s

Earth gravity g = 9.8 m/s

The maximum height reached by the bullet H_max = ?

Formula :

Maximum height H_max = (v₀²sin²θ)/2g

Substitute : v₀ =  410 m/s , θ = 29° and  g = 9.8 m/s.

H_max = (410²sin²29) / (2×9.8)

H_max = (410²×0.4848²) / (19.6)

H_max = 2015.83 m

Maximum Height  H_max = 2015.83 m

Answer 2

1.2.2)

Given data :

fired angle θ = 29°

initial velocity v₀ =  410 m/s

Earth gravity g = 9.8 m/s

The horizontal displacement reached by the bullet d = ?

Formula :

Horizontal displacementd = (v₀²sin2θ)/g

Substitute : v₀ =  410 m/s , θ = 29° and  g = 9.8 m/s.

d = (410²sin(2×29)) / (9.8)

d = (410²(0.848)) / (9.8)

d = 14546.62 m

The horizontal displacement is 14546.62 meters

Answer 3

1.3)

Speed of Vehicle V  v_v = 125 km/h in north-east direction.

Speed of Vehicle W  v_w = 125 km/h in east direction.

Consider that two vehicles are starts from same point.

By using vector analysis

OA = 125 km/h

OB = 125 km/h

OA = BC

OB = AC

AB = CD

OD = OB+BD

OC = Relative velocity magnitude

∠OCD = Relative velocity angle

From above figure

Angle ∠NOE = 90°

Angle ∠AOB is half in ∠NOE

Angle ∠AOB = ∠CBD = θ = 45°

From CBD right angled triangle

sinθ = BC/CD

BC = CD sin 45°

BC = 125 sin 45°

BC = 88.388 km/h

From CBD right angled triangle

cosθ = BD/CB

BD = CB cos 45°

BD = 125 cos 45°

BD = 88.388 km/h

OD = OB+BD = 125+88.388 = 213.388 km/h

From OAB right angled triangle

sinθ = AB/OA

AB = OA sin45

AB = 125 sin45

AB = 88.388 km/h

AB = CD = 88.388 km/h

From OCD right angled triangle

OC² = OD² + CD²

OC² = 213.388² + 88.388²

OC = 230.969 km/h

∠OCD = tan^-1(88.388/213.388)

∠OCD = 22.3°

Solution :

Magnitude of the velocity of vehicle W relative to the velocity of vehicle V is 230.969 km/h

Direction of the velocity of vehicle W relative to the velocity of vehicle V is 22.3° upside inclined to east