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A projectile is launched at an angle 'α' with the horizontal with a velocity 20 ms-1. After 10 s, its inclination with horizontal is 'β'. The value of tan β will be: (g = 10 ms-2)
- a)tan α + 5sec α
- b)tan α - 5sec α
- c)2tan α - 5sec α
- d)2tan α + 5sec α
Correct answer is option 'B'. Can you explain this answer?
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A projectile is launched at an angle α with the horizontal with ...
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A projectile is launched at an angle α with the horizontal with ...
Explanation:
Given data:
- Initial velocity of the projectile, u = 20 m/s
- Acceleration due to gravity, g = 10 m/s^2
- Time taken, t = 10 s
- Inclination with horizontal after 10 s, β
Concept:
- The horizontal component of velocity remains constant throughout the motion.
- The vertical component of velocity changes uniformly due to acceleration.
Calculation:
1. Horizontal component of velocity, u_x = u * cosα
2. Vertical component of velocity after 10 s, v_y = u * sinα - g * t
3. Tan β = v_y / u_x
4. Substitute the values from steps 1 and 2 into step 3:
Tan β = (u * sinα - g * t) / (u * cosα)
= (20 * sinα - 10 * 10) / (20 * cosα)
= 20 * sinα / 20 * cosα - 10
= tanα - 5
Therefore, the value of tan β is tanα - 5 secα. Hence, the correct answer is option 'b' (tanα - 5 secα).
Given data:
- Initial velocity of the projectile, u = 20 m/s
- Acceleration due to gravity, g = 10 m/s^2
- Time taken, t = 10 s
- Inclination with horizontal after 10 s, β
Concept:
- The horizontal component of velocity remains constant throughout the motion.
- The vertical component of velocity changes uniformly due to acceleration.
Calculation:
1. Horizontal component of velocity, u_x = u * cosα
2. Vertical component of velocity after 10 s, v_y = u * sinα - g * t
3. Tan β = v_y / u_x
4. Substitute the values from steps 1 and 2 into step 3:
Tan β = (u * sinα - g * t) / (u * cosα)
= (20 * sinα - 10 * 10) / (20 * cosα)
= 20 * sinα / 20 * cosα - 10
= tanα - 5
Therefore, the value of tan β is tanα - 5 secα. Hence, the correct answer is option 'b' (tanα - 5 secα).
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A projectile is launched at an angle α with the horizontal with a velocity 20 ms-1. After 10 s, its inclination with horizontal is β. The value of tan βwill be: (g = 10 ms-2)a)tanα + 5secαb)tanα - 5secαc)2tanα - 5secαd)2tanα + 5sec αCorrect answer is option 'B'. Can you explain this answer?
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