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A particle is thrown vertically up with an initial velocity 9 m/s form the surface of earth g = 10 m/s^2.the time taken by the particle to reach a height of 4 metre from the surface second time is?
Most Upvoted Answer
A particle is thrown vertically up with an initial velocity 9 m/s form...
U= 9 m/s. g=10m/s
by applying 2nd eqn of motion
s=ut +1/2 at^2
we get
s= ut - 1/2gt^2 as body is subjected upward ,it's going opposite to gravity.
we get,
5t^2-9t+4=0
by solving we get
t=1 or t=0.1 s
by applying 2nd eqn of motion
s=ut +1/2 at^2
we get
s= ut - 1/2gt^2 as body is subjected upward ,it's going opposite to gravity.
we get,
5t^2-9t+4=0
by solving we get
t=1 or t=0.1 s
Community Answer
A particle is thrown vertically up with an initial velocity 9 m/s form...
Given data:
Initial velocity (u) = 9 m/s
Acceleration due to gravity (g) = 10 m/s²
Height reached (h) = 4 m
Calculating time taken to reach the maximum height:
When the particle is thrown vertically up, it will reach the maximum height before falling down. At the maximum height, the particle's final velocity (v) will be zero.
Using the first equation of motion:
v = u + gt
At the maximum height, v = 0:
0 = 9 - 10t
10t = 9
t = 9/10
t = 0.9 s
So, it takes 0.9 seconds for the particle to reach the maximum height.
Calculating time taken to reach the height of 4 m for the second time:
To find the time taken to reach the height of 4 m again, we need to consider the total time taken for the particle to travel upwards and downwards.
1. Time taken to reach the maximum height (t₁) = 0.9 s (as calculated above)
2. Time taken to fall back from the maximum height (t₂) can be found using the second equation of motion:
h = ut + (1/2)gt²
4 = 0 + (1/2)(-10)t₂²
4 = -5t₂²
t₂² = -4/5
t₂ = √(-4/5)
Since time cannot be negative, t₂ will be imaginary. Therefore, the particle will not reach the height of 4 m again.
Summary:
- The time taken for the particle to reach the maximum height is 0.9 seconds.
- However, the particle will not reach the height of 4 m again as the time taken to fall back from the maximum height is imaginary.
Initial velocity (u) = 9 m/s
Acceleration due to gravity (g) = 10 m/s²
Height reached (h) = 4 m
Calculating time taken to reach the maximum height:
When the particle is thrown vertically up, it will reach the maximum height before falling down. At the maximum height, the particle's final velocity (v) will be zero.
Using the first equation of motion:
v = u + gt
At the maximum height, v = 0:
0 = 9 - 10t
10t = 9
t = 9/10
t = 0.9 s
So, it takes 0.9 seconds for the particle to reach the maximum height.
Calculating time taken to reach the height of 4 m for the second time:
To find the time taken to reach the height of 4 m again, we need to consider the total time taken for the particle to travel upwards and downwards.
1. Time taken to reach the maximum height (t₁) = 0.9 s (as calculated above)
2. Time taken to fall back from the maximum height (t₂) can be found using the second equation of motion:
h = ut + (1/2)gt²
4 = 0 + (1/2)(-10)t₂²
4 = -5t₂²
t₂² = -4/5
t₂ = √(-4/5)
Since time cannot be negative, t₂ will be imaginary. Therefore, the particle will not reach the height of 4 m again.
Summary:
- The time taken for the particle to reach the maximum height is 0.9 seconds.
- However, the particle will not reach the height of 4 m again as the time taken to fall back from the maximum height is imaginary.
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A particle is thrown vertically up with an initial velocity 9 m/s form the surface of earth g = 10 m/s^2.the time taken by the particle to reach a height of 4 metre from the surface second time is?
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A particle is thrown vertically up with an initial velocity 9 m/s form the surface of earth g = 10 m/s^2.the time taken by the particle to reach a height of 4 metre from the surface second time is? 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 thrown vertically up with an initial velocity 9 m/s form the surface of earth g = 10 m/s^2.the time taken by the particle to reach a height of 4 metre from the surface second time is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle is thrown vertically up with an initial velocity 9 m/s form the surface of earth g = 10 m/s^2.the time taken by the particle to reach a height of 4 metre from the surface second time is?.
A particle is thrown vertically up with an initial velocity 9 m/s form the surface of earth g = 10 m/s^2.the time taken by the particle to reach a height of 4 metre from the surface second time is? 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 thrown vertically up with an initial velocity 9 m/s form the surface of earth g = 10 m/s^2.the time taken by the particle to reach a height of 4 metre from the surface second time is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle is thrown vertically up with an initial velocity 9 m/s form the surface of earth g = 10 m/s^2.the time taken by the particle to reach a height of 4 metre from the surface second time is?.
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