Étiquette : JEE PHYSICS
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Motion Under Gravity: Key Concepts and Calculations
Question: A ball thrown up from the ground reaches a maximum height of 20 m. Find Solution: (a) Its initial velocity $${u=\sqrt{2gh}}$$ $${u=\sqrt {(2)(10)(20)}}$$ $${u=\sqrt 400 =20 \frac{m}{s}}$$ Its initial velocity is 20 ms-1 (b). The time taken to reach the maximum height $${u=gt}$$ $${t=\frac{u}{g}}$$ $${t=\frac {20}{10}}$$ $${t=2 s}$$ The time taken to reach maximum…
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Motion Under Gravity Numerical: Calculating Ball’s Speed at Point C
Question: A ball is thrown vertically upwards with an initial velocity of 5 ms-1 from point A as shown below. C is a point 10 m vertically below the point A. The the speed of the ball at point C will be(take g=10 ms-2 and neglect air resistance) Solution: Initial velocity u= +5 ms-1 (From…
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Building Height Calculation Using Motion Equations
Question: A particle is projected up with an initial speed of u=10 ms-1 from the top of a building at time t=0.At time t=5 sec, the particle strikes the ground.Find the height of the building. Solution:Building height calculation Solution: Net displacement of the particle is the difference of final position(C) and the initial position(A). Applying…
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Vertical Projectile Motion: Height and Time Calculations
Question: A ball is thrown vertically upwards with a velocity of 20 ms-1 from the top of a multistorey building . The height of the point from where the ball is thrown is 25.0 m from the ground.(a) How high will the ball rise? and (b) How long will it be before the ball hits…
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A ball is thrown upwards from the ground with an initial velocity ‘u’.The ball is at a height of 60 m at two times,the interval being 6 s.Find the initial velocity ‘u’
Solution: First Method: The ball reaches the given height of 60 m at a time t 1 seconds ( rising up).For the second time it reaches the 60 m height at a time t 2 seconds. (comes down) Applying the equation of motion with the sign convention. Initial velocity (u) = +u Acceleration due to…
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