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A person standing on a tower of height 60 m throws an object upwards with velocity of 40 m/s at an angle 30° to horizontal. Find the total time taken by the object to gain maximum height and fall on the ground (take g = 10 m/s2)
- a)3 s
- b)20 s
- c)6 s
- d)16 s
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A person standing on a tower of height 60 m throws an object upwards w...
Projectile motion:
- It is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
- Initial Velocity: The initial velocity can be given as x components and y components.
- Component of initial velocity in x-direction, (ux) = u cos θ
- Component of initial velocity in the y-direction, (uy) = u sin θ
- In the case of projectile motion, we can see a free-fall motion of a body on a parabolic path with constant velocity.
- If a body is thrown at a certain angle then during its movement, we get two components of velocity as given below.
- And thus, the range of a projectile is the displacement of a particle along the x-axis and can be given as:
The height of projectile is given by,
- Whereas the time of flight is the total time for which projectile stayed in the air.
Time of flight for the projectile,
where The angle of projection = θ, Initial velocity = u, Gravitational acceleration = g, Time of flight = t, Range of projectile = R
Calculation:
Given:
Height of the tower H1 = 60 m, initial velocity of the object u = 40 m/s, angle of inclination with horizontal θ = 30°, ux = 40 × cos 30° = 20√3 m/s, uy = 40 × sin 30° = 20 m/s
Time taken to reach the maximum height is given by,
Height above the tower is given by,
Therefore, the maximum height is, Hmax = H1 + H2 = 60 + 20 = 80 m
Now, the time taken to free fall from maximum height is,
Thus, the total time is taken during the entire flight is given by, Ttotal = T1 + T2 = 2 + 4 = 6 s
where The angle of projection = θ, Initial velocity = u, Gravitational acceleration = g, Time of flight = t, Range of projectile = R
Calculation:
Given:
Height of the tower H1 = 60 m, initial velocity of the object u = 40 m/s, angle of inclination with horizontal θ = 30°, ux = 40 × cos 30° = 20√3 m/s, uy = 40 × sin 30° = 20 m/s
Time taken to reach the maximum height is given by,
Height above the tower is given by,
Therefore, the maximum height is, Hmax = H1 + H2 = 60 + 20 = 80 m
Now, the time taken to free fall from maximum height is,
Thus, the total time is taken during the entire flight is given by, Ttotal = T1 + T2 = 2 + 4 = 6 s
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A person standing on a tower of height 60 m throws an object upwards w...
To solve this problem, we can break down the given information into components and use the equations of motion.
First, let's find the initial vertical velocity (Vy) and initial horizontal velocity (Vx) of the object.
Given:
Initial velocity (Vo) = 40 m/s
Angle (θ) = 30°
Using trigonometry, we can find the vertical and horizontal components of the initial velocity:
Vy = Vo * sin(θ)
Vx = Vo * cos(θ)
Vy = 40 * sin(30°)
Vy = 40 * 0.5
Vy = 20 m/s
Vx = 40 * cos(30°)
Vx = 40 * √3/2
Vx = 20√3 m/s
Now, let's consider the vertical motion of the object.
Given:
Initial vertical velocity (Vy) = 20 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Final vertical velocity (Vf) = 0 m/s (at the peak)
We can use the equation of motion to find the time taken to reach the peak of the trajectory:
Vf = Vy + gt
0 = 20 + (-9.8)t
-20 = -9.8t
t = -20 / -9.8
t ≈ 2.04 seconds
The object takes approximately 2.04 seconds to reach the peak of its trajectory.
Now, let's consider the horizontal motion of the object.
Given:
Initial horizontal velocity (Vx) = 20√3 m/s
Distance (d) = ?
We can use the equation of motion to find the horizontal distance traveled by the object:
d = Vx * t
d ≈ (20√3 m/s) * (2.04 seconds)
d ≈ 40√3 * 2.04 m
d ≈ 81.6√3 m
Therefore, the horizontal distance traveled by the object is approximately 81.6√3 meters.
In conclusion, when a person standing on a tower of height 60 m throws an object upwards with a velocity of 40 m/s at an angle of 30°, the object reaches a peak height of 20 meters and travels approximately 81.6√3 meters horizontally.
First, let's find the initial vertical velocity (Vy) and initial horizontal velocity (Vx) of the object.
Given:
Initial velocity (Vo) = 40 m/s
Angle (θ) = 30°
Using trigonometry, we can find the vertical and horizontal components of the initial velocity:
Vy = Vo * sin(θ)
Vx = Vo * cos(θ)
Vy = 40 * sin(30°)
Vy = 40 * 0.5
Vy = 20 m/s
Vx = 40 * cos(30°)
Vx = 40 * √3/2
Vx = 20√3 m/s
Now, let's consider the vertical motion of the object.
Given:
Initial vertical velocity (Vy) = 20 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Final vertical velocity (Vf) = 0 m/s (at the peak)
We can use the equation of motion to find the time taken to reach the peak of the trajectory:
Vf = Vy + gt
0 = 20 + (-9.8)t
-20 = -9.8t
t = -20 / -9.8
t ≈ 2.04 seconds
The object takes approximately 2.04 seconds to reach the peak of its trajectory.
Now, let's consider the horizontal motion of the object.
Given:
Initial horizontal velocity (Vx) = 20√3 m/s
Distance (d) = ?
We can use the equation of motion to find the horizontal distance traveled by the object:
d = Vx * t
d ≈ (20√3 m/s) * (2.04 seconds)
d ≈ 40√3 * 2.04 m
d ≈ 81.6√3 m
Therefore, the horizontal distance traveled by the object is approximately 81.6√3 meters.
In conclusion, when a person standing on a tower of height 60 m throws an object upwards with a velocity of 40 m/s at an angle of 30°, the object reaches a peak height of 20 meters and travels approximately 81.6√3 meters horizontally.
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A person standing on a tower of height 60 m throws an object upwards with velocity of 40 m/s at an angle 30° to horizontal. Find the total time taken by the object to gain maximum height and fall on the ground (take g = 10 m/s2)a)3 sb)20 sc)6 sd)16 sCorrect answer is option 'C'. Can you explain this answer?
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A person standing on a tower of height 60 m throws an object upwards with velocity of 40 m/s at an angle 30° to horizontal. Find the total time taken by the object to gain maximum height and fall on the ground (take g = 10 m/s2)a)3 sb)20 sc)6 sd)16 sCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A person standing on a tower of height 60 m throws an object upwards with velocity of 40 m/s at an angle 30° to horizontal. Find the total time taken by the object to gain maximum height and fall on the ground (take g = 10 m/s2)a)3 sb)20 sc)6 sd)16 sCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A person standing on a tower of height 60 m throws an object upwards with velocity of 40 m/s at an angle 30° to horizontal. Find the total time taken by the object to gain maximum height and fall on the ground (take g = 10 m/s2)a)3 sb)20 sc)6 sd)16 sCorrect answer is option 'C'. Can you explain this answer?.
A person standing on a tower of height 60 m throws an object upwards with velocity of 40 m/s at an angle 30° to horizontal. Find the total time taken by the object to gain maximum height and fall on the ground (take g = 10 m/s2)a)3 sb)20 sc)6 sd)16 sCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A person standing on a tower of height 60 m throws an object upwards with velocity of 40 m/s at an angle 30° to horizontal. Find the total time taken by the object to gain maximum height and fall on the ground (take g = 10 m/s2)a)3 sb)20 sc)6 sd)16 sCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A person standing on a tower of height 60 m throws an object upwards with velocity of 40 m/s at an angle 30° to horizontal. Find the total time taken by the object to gain maximum height and fall on the ground (take g = 10 m/s2)a)3 sb)20 sc)6 sd)16 sCorrect answer is option 'C'. Can you explain this answer?.
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