Read More A projectile is projected with initial velocity of 30 m/s at an angle =tan-¹ (¾)
After 1 second, direction of motion a of the particle makes an angle a with horizontal, then angle a is given as (g = 10 m s-2) 1 tan
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Appropriate Question
A particle is projected with velocity of 30m/s (initial velocity) at an angle of θ = tanˉ¹ (3/4). After 1 second, direction of motion a particle makes an angle 'α' with horizontal, then angle 'α' is given by (g = 10 m/s²).
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★ Concept
Here, the concept of Projectile motion has been used. Let's have a look over provided conditions first, it's given that tanθ = 3/4, now using pythagoras theoram, we will find cosθ () and sinθ () i.e, horizontal distance and vertical distance velocity respectively. Horizontal distance velocity will remain constant since no acc. is acting on a particle horizontally, but vertical distance velocity will be reduced at a rate of 10 msˉ¹ per second. Now, using the concept that free vectors can be moved parallel we'll find the value for tan 'α'.
Let's proceed with Calculation !!
Given that,
θ = tanˉ¹ (3/4) ; tan θ = 3/4.
[tanA = P/B, where 'A', refers angle]
P = 3K ; B = 4K ; H = ? (By Pythagoras Theorem)
H² = P² + B² = (3K)² + (4K)² = 25K²
H = 5K
cosθ = 4/5 ; sinθ = 3/5
Now, Finding Horizontal and Vertical velocity
Horizontal Distance (Ux)
Vertical Distance (Uy)
Horizontal distance velocity will remain constant since no acceleration is acting on a particle horizontally so, vcos α = 24 msˉ¹ __(1)
For Vertical Distance Velocity
Newton's First equation of Motion
(v = u + at )
Either By the Concept of Moving Free Vectors Parallel or by dividing eq (2) by (1)
Tan α = vsinα/vcosα = 8/24 = 1/3
∴ Tan α = 1/3 , required angle.
➺ Additional Information
1) The path followed under projectile motion is known as trajectory.
2) The velocity of the particle undergoing projectile motion at any point on its path is directed along the tangent at that point.
3) If a Vector is resolved into two vectors whose combined effect is the same as that of the given vector, then the resolved vectors are called the 'components' of the given vector. For e.g- Velocity is itself resolved into horizontal and vertical component. This is known as 'Resolution of Vectors'.
Answer:
Appropriate Question
A particle is projected with velocity of 30m/s (initial velocity) at an angle of θ = tanˉ¹ (3/4). After 1 second, direction of motion a particle makes an angle 'α' with horizontal, then angle 'α' is given by (g = 10 m/s²).
_____________________________________
★ Concept
Here, the concept of Projectile motion has been used. Let's have a look over provided conditions first, it's given that tanθ = 3/4, now using pythagoras theoram, we will find cosθ
=usinθ ) i.e, horizontal distance and vertical distance velocity respectively. Horizontal distance velocity will remain constant since no acc. is acting on a particle horizontally, but vertical distance velocity will be reduced at a rate of 10 msˉ¹ per second. Now, using the concept that free vectors can be moved parallel we'll find the value for tan 'α'.
Let's proceed with Calculation !!
Given that,
θ = tanˉ¹ (3/4) ; tan θ = 3/4.
[tanA = P/B, where 'A', refers angle]
P = 3K ; B = 4K ; H = ? (By Pythagoras Theorem)
H² = P² + B² = (3K)² + (4K)² = 25K²
H = 5K
cosθ = 4/5 ; sinθ = 3/5
Now, Finding Horizontal and Vertical velocity
Horizontal Distance (Ux)
Vertical Distance (Uy)
Horizontal distance velocity will remain constant since no acceleration is acting on a particle horizontally so, vcos α = 24 msˉ¹ __(1)
For Vertical Distance Velocity
Newton's First equation of Motion
(v = u + at )
Either By the Concept of Moving Free Vectors Parallel or by dividing eq (2) by (1)
Tan α = vsinα/vcosα = 8/24 = 1/3
∴ Tan α = 1/3 , required angle.
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