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If the velocity of the part projectile is b bar equals to 20 ICAP + 10 J cap metre per second when it is at a height of 15 m above the ground then the then based on this information and taking g=10m/s (i) The Launch Speed in m/s is (ii) The coordinate where the maximum height is attained is?
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If the velocity of the part projectile is b bar equals to 20 ICAP + 10...
Launch Speed:
- Given velocity vector b = 20 î + 10 ĵ m/s
- Launch speed (magnitude of velocity) = √(20^2 + 10^2) = √(400 + 100) = √500 = 22.36 m/s
Coordinate of Maximum Height:
- At the maximum height, the vertical component of velocity becomes zero (V_y = 0)
- Using the equation of motion: V_f = V_i + at, where V_f = 0, V_i = 10 m/s (initial vertical velocity), a = -10 m/s^2 (acceleration due to gravity)
- 0 = 10 - 10t
- Solving for t, we get t = 1 second
- Substitute t = 1 second into the equation: h = V_i * t + (1/2) * a * t^2
- h = 10 * 1 + (1/2) * (-10) * 1^2 = 10 - 5 = 5 m
- Therefore, the maximum height is attained at a height of 5 meters above the ground.
- Given velocity vector b = 20 î + 10 ĵ m/s
- Launch speed (magnitude of velocity) = √(20^2 + 10^2) = √(400 + 100) = √500 = 22.36 m/s
Coordinate of Maximum Height:
- At the maximum height, the vertical component of velocity becomes zero (V_y = 0)
- Using the equation of motion: V_f = V_i + at, where V_f = 0, V_i = 10 m/s (initial vertical velocity), a = -10 m/s^2 (acceleration due to gravity)
- 0 = 10 - 10t
- Solving for t, we get t = 1 second
- Substitute t = 1 second into the equation: h = V_i * t + (1/2) * a * t^2
- h = 10 * 1 + (1/2) * (-10) * 1^2 = 10 - 5 = 5 m
- Therefore, the maximum height is attained at a height of 5 meters above the ground.
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If the velocity of the part projectile is b bar equals to 20 ICAP + 10 J cap metre per second when it is at a height of 15 m above the ground then the then based on this information and taking g=10m/s (i) The Launch Speed in m/s is (ii) The coordinate where the maximum height is attained is?
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If the velocity of the part projectile is b bar equals to 20 ICAP + 10 J cap metre per second when it is at a height of 15 m above the ground then the then based on this information and taking g=10m/s (i) The Launch Speed in m/s is (ii) The coordinate where the maximum height is attained is? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about If the velocity of the part projectile is b bar equals to 20 ICAP + 10 J cap metre per second when it is at a height of 15 m above the ground then the then based on this information and taking g=10m/s (i) The Launch Speed in m/s is (ii) The coordinate where the maximum height is attained is? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the velocity of the part projectile is b bar equals to 20 ICAP + 10 J cap metre per second when it is at a height of 15 m above the ground then the then based on this information and taking g=10m/s (i) The Launch Speed in m/s is (ii) The coordinate where the maximum height is attained is?.
If the velocity of the part projectile is b bar equals to 20 ICAP + 10 J cap metre per second when it is at a height of 15 m above the ground then the then based on this information and taking g=10m/s (i) The Launch Speed in m/s is (ii) The coordinate where the maximum height is attained is? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about If the velocity of the part projectile is b bar equals to 20 ICAP + 10 J cap metre per second when it is at a height of 15 m above the ground then the then based on this information and taking g=10m/s (i) The Launch Speed in m/s is (ii) The coordinate where the maximum height is attained is? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the velocity of the part projectile is b bar equals to 20 ICAP + 10 J cap metre per second when it is at a height of 15 m above the ground then the then based on this information and taking g=10m/s (i) The Launch Speed in m/s is (ii) The coordinate where the maximum height is attained is?.
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