A cruise missile under test is moving horizontally at Ma =2 in the atmosphere at an elevation of 2000 m (Air temperature is 2 °C
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Answer:
Explanation:
<u>Projectile Motion</u>
In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:
Where is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.
The horizontal and vertical distances are, respectively:
The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find
Using this time in the horizontal distance, we find the Range or maximum horizontal distance:
Let's solve for
This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:
Or equivalently:
Given Vo=37 m/s and R=70 m
And
Answer:
Magnitude of the velocity = 16.82 m/s
Angular direction, θ = 52.41°
Explanation:
As given ,
mass of bullet, m₁= 50g = 0.05 kg
speed of bullet , u₁ = 600 m/s
mass of the block , m₂ = 4 kg
speed of the block before collision , u₂ = 12 m/s
direction , θ = 30°
Now,
Assume that the combined velocity of bullet and block after collision = v
and the direction = θ
Now, from the conservation of momentum in x - direction :
m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) vₓ
where v = final velocity after collision
u₁ = initial velocity of bullet before collision = 0
m₁ = mass of the bullet before collision = 0.05 kg
u₂ = velocity of block before collision = 12 cos(30° )
m₂ = mass of block before collision
m₁ + m₂ = combined mass of bullet and block after collision = 0.05 + 4
∴ we get
0.05 (0) + 4(12 cos(30° ) ) = ( 0.05 + 4 ) vₓ
⇒ 0 + 4(6√3) = 4.05 vₓ
⇒24√3 = 4.05 vₓ
⇒vₓ = 10.26 m/s
Now, from the conservation of momentum in y - direction :
m₁ u₁ + m₂ u₂ = ( m₁ + m₂ )
where v = final velocity after collision
u₁= initial velocity of bullet before collision = 600
m₁ = mass of the bullet before collision = 0.05 kg
u₂ = velocity of block before collision = 12 sin(30° )
m₂= mass of block before collision
m₁+ m₂= combined mass of bullet and block after collision = 0.05 + 4
∴ we get
0.05 (600) + 4(12 sin(30° ) ) = ( 0.05 + 4 )
⇒ 30 + 4(6) = 4.05
⇒30 +24 = 4.05
⇒54 = 4.05
⇒ = 13.33 m/s
Now, the magnitude of the velocity = √vₓ² + ² = √(10.26)² + (13.33)²
= √105.26 + 177.68
= √282.95 = 16.82
The angular direction, θ = (
) =
(
) =
(1.299) = 52.41°