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A particle of mass 50 g is projected horizontally Y from the top of a tower of height 80 m with a velocity 40 m/s. If g = 10 * m / s ^ 2 then find the :- (take downward as positive y-direction) (1) Velocity at the instant t = 3s (2) Position vector at the instant t = 2 s (3) Velocity just before hitting the ground (4) Change in velocity in 2 s?
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A particle of mass 50 g is projected horizontally Y from the top of a ...
Given Information:
Mass of the particle (m) = 50 g = 0.05 kg
Height of the tower (h) = 80 m
Initial velocity (u) = 40 m/s
Acceleration due to gravity (g) = 10 m/s^2
Answer:
1) Velocity at the instant t = 3s:
The particle is projected horizontally, so there is no vertical component of velocity. The horizontal velocity remains constant throughout the motion.
Horizontal Velocity:
The horizontal velocity (Vx) remains constant and equal to the initial horizontal velocity (u).
Vx = u = 40 m/s
Vertical Velocity:
The vertical velocity (Vy) changes due to the acceleration due to gravity.
Vy = u + gt
= 0 + (10 m/s^2)(3 s)
= 30 m/s
Resultant Velocity:
The resultant velocity (V) is the vector sum of the horizontal and vertical velocities.
V = √(Vx^2 + Vy^2)
= √((40 m/s)^2 + (30 m/s)^2)
= √(1600 + 900)
= √(2500)
= 50 m/s
Therefore, the velocity of the particle at t = 3 s is 50 m/s.
2) Position vector at the instant t = 2 s:
The position vector of the particle at a given time (t) is given by:
R = Ri + Vti + (1/2)gt^2
Where,
Ri = Initial position vector (top of the tower)
Vt = Initial velocity vector (horizontal velocity)
g = Acceleration due to gravity
t = Time
Initial Position:
The initial position of the particle is at the top of the tower. Let's assume this position as the origin of the coordinate system. Therefore, the initial position vector (Ri) is zero.
Ri = 0
Initial Velocity:
The initial velocity of the particle is the horizontal velocity (u).
Vt = u = 40 m/s
Acceleration due to Gravity:
The acceleration due to gravity acts vertically downward, so its vector representation is given by -g.
g = -10 m/s^2
Position Vector:
Substituting the values into the position vector equation:
R = Ri + Vti + (1/2)gt^2
= 0 + (40 m/s)(2 s) + (1/2)(-10 m/s^2)(2 s)^2
= 0 + 80 m + (-10 m/s^2)(4 s^2)
= 80 m - 40 m
= 40 m
Therefore, the position vector of the particle at t = 2 s is 40 m.
3) Velocity just before hitting the ground:
The particle falls freely under the influence of gravity after being projected horizontally. The time taken to reach the ground can be found using the equation:
h = (1/2)gt^2
Where,
h = Height of the tower
g = Acceleration due
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A particle of mass 50 g is projected horizontally Y from the top of a tower of height 80 m with a velocity 40 m/s. If g = 10 * m / s ^ 2 then find the :- (take downward as positive y-direction) (1) Velocity at the instant t = 3s (2) Position vector at the instant t = 2 s (3) Velocity just before hitting the ground (4) Change in velocity in 2 s?
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A particle of mass 50 g is projected horizontally Y from the top of a tower of height 80 m with a velocity 40 m/s. If g = 10 * m / s ^ 2 then find the :- (take downward as positive y-direction) (1) Velocity at the instant t = 3s (2) Position vector at the instant t = 2 s (3) Velocity just before hitting the ground (4) Change in velocity in 2 s? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle of mass 50 g is projected horizontally Y from the top of a tower of height 80 m with a velocity 40 m/s. If g = 10 * m / s ^ 2 then find the :- (take downward as positive y-direction) (1) Velocity at the instant t = 3s (2) Position vector at the instant t = 2 s (3) Velocity just before hitting the ground (4) Change in velocity in 2 s? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 50 g is projected horizontally Y from the top of a tower of height 80 m with a velocity 40 m/s. If g = 10 * m / s ^ 2 then find the :- (take downward as positive y-direction) (1) Velocity at the instant t = 3s (2) Position vector at the instant t = 2 s (3) Velocity just before hitting the ground (4) Change in velocity in 2 s?.
A particle of mass 50 g is projected horizontally Y from the top of a tower of height 80 m with a velocity 40 m/s. If g = 10 * m / s ^ 2 then find the :- (take downward as positive y-direction) (1) Velocity at the instant t = 3s (2) Position vector at the instant t = 2 s (3) Velocity just before hitting the ground (4) Change in velocity in 2 s? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle of mass 50 g is projected horizontally Y from the top of a tower of height 80 m with a velocity 40 m/s. If g = 10 * m / s ^ 2 then find the :- (take downward as positive y-direction) (1) Velocity at the instant t = 3s (2) Position vector at the instant t = 2 s (3) Velocity just before hitting the ground (4) Change in velocity in 2 s? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 50 g is projected horizontally Y from the top of a tower of height 80 m with a velocity 40 m/s. If g = 10 * m / s ^ 2 then find the :- (take downward as positive y-direction) (1) Velocity at the instant t = 3s (2) Position vector at the instant t = 2 s (3) Velocity just before hitting the ground (4) Change in velocity in 2 s?.
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