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A particle is projected vertically up from the top of a tower with velocity 10 m/sec. It reaches the ground in 5 sec. Find, (i) Height of the tower, (ii) Striking velocity of particle at ground, (iii) Distance traversed by the particle and Average speed and average velocity of particle?
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Problem: A particle is projected vertically up from the top of a tower with velocity 10 m/sec. It reaches the ground in 5 sec. Find, (i) Height of the tower, (ii) Striking velocity of particle at ground, (iii) Distance traversed by the particle and Average speed and average velocity of particle?
Solution:
(i) Height of the tower:
Let H be the height of the tower. Using the formula, H= ut + 1/2at^2, where u= initial velocity, a= acceleration due to gravity and t= time taken to reach the ground, we get,
H = ut + 1/2at^2
= 10*5 + 1/2*9.8*5^2
= 250 m
Therefore, the height of the tower is 250 m.
(ii) Striking velocity of particle at ground:
Let V be the striking velocity of the particle at the ground. Using the formula, V = u + at, where u= initial velocity, a= acceleration due to gravity and t= time taken to reach the ground, we get,
V = u + at
= 10 + 9.8*5
= 59 m/s
Therefore, the striking velocity of the particle at the ground is 59 m/s.
(iii) Distance traversed by the particle and Average speed and average velocity of particle:
Let S be the distance traversed by the particle. Using the formula, S= ut + 1/2at^2, where u= initial velocity, a= acceleration due to gravity and t= time taken to reach the ground, we get,
S = ut + 1/2at^2
= 10*5 - 1/2*9.8*5^2
= 125 m
Therefore, the distance traversed by the particle is 125 m.
The average speed of the particle is given by,
Average speed = Total distance / Total time taken
= (125 + 250) / 5
= 75 m/s
The average velocity of the particle is zero since it moves upward with a velocity of 10 m/s and then downwards with a velocity of 59 m/s. Therefore, the displacement of the particle is zero and the average velocity is zero.
Conclusion: In this problem, we found the height of the tower to be 250 m, striking velocity of the particle at the ground to be 59 m/s, distance traversed by the particle to be 125 m, average speed of the particle to be 75 m/s and average velocity to be zero.
Solution:
(i) Height of the tower:
Let H be the height of the tower. Using the formula, H= ut + 1/2at^2, where u= initial velocity, a= acceleration due to gravity and t= time taken to reach the ground, we get,
H = ut + 1/2at^2
= 10*5 + 1/2*9.8*5^2
= 250 m
Therefore, the height of the tower is 250 m.
(ii) Striking velocity of particle at ground:
Let V be the striking velocity of the particle at the ground. Using the formula, V = u + at, where u= initial velocity, a= acceleration due to gravity and t= time taken to reach the ground, we get,
V = u + at
= 10 + 9.8*5
= 59 m/s
Therefore, the striking velocity of the particle at the ground is 59 m/s.
(iii) Distance traversed by the particle and Average speed and average velocity of particle:
Let S be the distance traversed by the particle. Using the formula, S= ut + 1/2at^2, where u= initial velocity, a= acceleration due to gravity and t= time taken to reach the ground, we get,
S = ut + 1/2at^2
= 10*5 - 1/2*9.8*5^2
= 125 m
Therefore, the distance traversed by the particle is 125 m.
The average speed of the particle is given by,
Average speed = Total distance / Total time taken
= (125 + 250) / 5
= 75 m/s
The average velocity of the particle is zero since it moves upward with a velocity of 10 m/s and then downwards with a velocity of 59 m/s. Therefore, the displacement of the particle is zero and the average velocity is zero.
Conclusion: In this problem, we found the height of the tower to be 250 m, striking velocity of the particle at the ground to be 59 m/s, distance traversed by the particle to be 125 m, average speed of the particle to be 75 m/s and average velocity to be zero.
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A particle is projected vertically up from the top of a tower with velocity 10 m/sec. It reaches the ground in 5 sec. Find, (i) Height of the tower, (ii) Striking velocity of particle at ground, (iii) Distance traversed by the particle and Average speed and average velocity of particle?
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A particle is projected vertically up from the top of a tower with velocity 10 m/sec. It reaches the ground in 5 sec. Find, (i) Height of the tower, (ii) Striking velocity of particle at ground, (iii) Distance traversed by the particle and Average speed and average velocity of particle? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A particle is projected vertically up from the top of a tower with velocity 10 m/sec. It reaches the ground in 5 sec. Find, (i) Height of the tower, (ii) Striking velocity of particle at ground, (iii) Distance traversed by the particle and Average speed and average velocity of particle? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle is projected vertically up from the top of a tower with velocity 10 m/sec. It reaches the ground in 5 sec. Find, (i) Height of the tower, (ii) Striking velocity of particle at ground, (iii) Distance traversed by the particle and Average speed and average velocity of particle?.
A particle is projected vertically up from the top of a tower with velocity 10 m/sec. It reaches the ground in 5 sec. Find, (i) Height of the tower, (ii) Striking velocity of particle at ground, (iii) Distance traversed by the particle and Average speed and average velocity of particle? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A particle is projected vertically up from the top of a tower with velocity 10 m/sec. It reaches the ground in 5 sec. Find, (i) Height of the tower, (ii) Striking velocity of particle at ground, (iii) Distance traversed by the particle and Average speed and average velocity of particle? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle is projected vertically up from the top of a tower with velocity 10 m/sec. It reaches the ground in 5 sec. Find, (i) Height of the tower, (ii) Striking velocity of particle at ground, (iii) Distance traversed by the particle and Average speed and average velocity of particle?.
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