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A projectile is thrown in the upward direction making an angle of 60º with the horizontal direction, with a velocity of 147 ms–1. Then the time after which inclination with the horizontal is 45º, is
- a)15 (√3 - 1) s
- b)15 (√3 + 1) s
- c)7.5 (√3 - 1) s
- d)7.5 (√3 + 1) s
Correct answer is option 'C'. Can you explain this answer?
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A projectile is thrown in the upward direction making an angle of 60º...
Given data:
- Initial velocity, u = 147 m/s
- Angle made with horizontal, θ1 = 60°
- Angle made with horizontal after some time, θ2 = 45°
We need to find the time taken to reach the inclination of 45°.
Approach:
We can use the equations of motion to solve this problem. Let's break the initial velocity into its horizontal and vertical components.
Initial velocity, u = 147 m/s
Horizontal component, u_x = u cos θ1 = 147 × cos 60° = 73.5 m/s
Vertical component, u_y = u sin θ1 = 147 × sin 60° = 127.3 m/s
Now, let's find the time taken to reach the maximum height. At the highest point, the vertical component of velocity becomes zero.
Final velocity, v = 0 m/s (at highest point)
Vertical component of final velocity, v_y = 0 m/s
Acceleration due to gravity, a = 9.8 m/s^2
Using the equation v_y = u_y + at, we can find the time taken to reach the maximum height.
t = (v_y - u_y) / a
t = (0 - 127.3) / (-9.8)
t = 12.98 s (approx)
Now, let's find the time taken to reach the inclination of 45° from the highest point.
Using the equation θ = tan^-1 (v_y / v_x), we can find the angle made with the horizontal.
For the highest point, θ = 90°, so v_x = u_x and v_y = 0.
θ = tan^-1 (0 / 73.5)
θ = 0°
Now, let's find the time taken to reach the inclination of 45°.
θ = 45°
v_x = v cos θ = v sin θ = (u sin θ1) / 2 = 73.5 / √2 = 51.96 m/s
v_y = v sin θ = (u cos θ1) / 2 = 127.3 / √2 = 89.85 m/s
Using the equation v_y = u_y + at, we can find the time taken to reach the inclination of 45°.
v_y = u_y + at
t = (v_y - u_y) / a
t = (89.85 - (-127.3)) / 9.8
t = 7.5 (√3 - 1) s
Therefore, option C, 7.5 (√3 - 1) s, is the correct answer.
- Initial velocity, u = 147 m/s
- Angle made with horizontal, θ1 = 60°
- Angle made with horizontal after some time, θ2 = 45°
We need to find the time taken to reach the inclination of 45°.
Approach:
We can use the equations of motion to solve this problem. Let's break the initial velocity into its horizontal and vertical components.
Initial velocity, u = 147 m/s
Horizontal component, u_x = u cos θ1 = 147 × cos 60° = 73.5 m/s
Vertical component, u_y = u sin θ1 = 147 × sin 60° = 127.3 m/s
Now, let's find the time taken to reach the maximum height. At the highest point, the vertical component of velocity becomes zero.
Final velocity, v = 0 m/s (at highest point)
Vertical component of final velocity, v_y = 0 m/s
Acceleration due to gravity, a = 9.8 m/s^2
Using the equation v_y = u_y + at, we can find the time taken to reach the maximum height.
t = (v_y - u_y) / a
t = (0 - 127.3) / (-9.8)
t = 12.98 s (approx)
Now, let's find the time taken to reach the inclination of 45° from the highest point.
Using the equation θ = tan^-1 (v_y / v_x), we can find the angle made with the horizontal.
For the highest point, θ = 90°, so v_x = u_x and v_y = 0.
θ = tan^-1 (0 / 73.5)
θ = 0°
Now, let's find the time taken to reach the inclination of 45°.
θ = 45°
v_x = v cos θ = v sin θ = (u sin θ1) / 2 = 73.5 / √2 = 51.96 m/s
v_y = v sin θ = (u cos θ1) / 2 = 127.3 / √2 = 89.85 m/s
Using the equation v_y = u_y + at, we can find the time taken to reach the inclination of 45°.
v_y = u_y + at
t = (v_y - u_y) / a
t = (89.85 - (-127.3)) / 9.8
t = 7.5 (√3 - 1) s
Therefore, option C, 7.5 (√3 - 1) s, is the correct answer.
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A projectile is thrown in the upward direction making an angle of 60º with the horizontal direction, with a velocity of 147 ms–1. Then the time after which inclination with the horizontal is 45º, isa)15 (√3 - 1) sb)15 (√3 + 1) sc)7.5 (√3 - 1) sd)7.5 (√3 + 1) sCorrect answer is option 'C'. Can you explain this answer?
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A projectile is thrown in the upward direction making an angle of 60º with the horizontal direction, with a velocity of 147 ms–1. Then the time after which inclination with the horizontal is 45º, isa)15 (√3 - 1) sb)15 (√3 + 1) sc)7.5 (√3 - 1) sd)7.5 (√3 + 1) sCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A projectile is thrown in the upward direction making an angle of 60º with the horizontal direction, with a velocity of 147 ms–1. Then the time after which inclination with the horizontal is 45º, isa)15 (√3 - 1) sb)15 (√3 + 1) sc)7.5 (√3 - 1) sd)7.5 (√3 + 1) sCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A projectile is thrown in the upward direction making an angle of 60º with the horizontal direction, with a velocity of 147 ms–1. Then the time after which inclination with the horizontal is 45º, isa)15 (√3 - 1) sb)15 (√3 + 1) sc)7.5 (√3 - 1) sd)7.5 (√3 + 1) sCorrect answer is option 'C'. Can you explain this answer?.
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