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If a ball is thrown vertically upwards with speed `u`, the distance covered during the last `t` second of its ascent is

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A body is projected vertically upward with speed 40m/s. The distance travelled by body in the last second of upward journey is [take g=9.8m/s^(2) and neglect eggect of air rresistance]:-

From a tower of height H , a particle is thrown vertically upwards with a speed u . The time taken by the particle , to hit the ground , is n times that taken by it to reach the highest point of its path. The relative between H, u and n is :

A body is projected vertically up. What is the distance covered in its last second of upward motion? (g=10 m//s^(2))

A ball is thrown vertiocally upwards with some speed. It reaches two points A and B one after another such that heights of A and B are one fourth and three-fourth of the maximum height attained. If the total time of flight if T. the maximum time taken by the ball to travel from A and B, is:-

A ball is dropped from the roof of a tower height h . The total distance covered by it in the last second of its motion is equal to the distance covered by it in first three seconds. The value of h in metre is (g=10m//s^(2))

A ball is thrown downwards with a speed of 20 ms^(–1) from top of a building 150m high and simultaneously another ball is thrown vertically upwards with a speed of 30 ms^(–1) from the foot of the building. Find the time after which both the balls will meet - (g = 10 ms^(–2))

A particle is thrown vertically upward with a speed u from the top of a tower of height h from ground level, If after first impact with ground it just reaches to height h from ground the coeffecient of restitution for the collision is :-

A ball is thrown vertically upwards with a velocity v and an initial kinetic energy E . When half way to the top of its flight, it can have velocity and kinetic energy respectively of :

Assertion: If a body is dropped from the top of a twoer of height h and another body is thrown up simultaneously with velocity u from the foot of the tower, then both of them would meet after a time (h)/(u) . Reason: For a body projected upwards, the distance covered by the body in the last second of its upwards journey is always 4.9m irrespective of velocity of projection (g=9.8 m//s^(2))

If a ball is thrown vertically upwards with a velocity of 40m//s , then velocity of the ball after 2s will be (g = 10m//s^(2))

ALLEN -MOTION IN A PALNE-EXERCISE-02
  1. Three different objects of masses m(1) , m(2) and m(2) are allowed to ...

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  2. A ball is thrown upwards . Its height varies with time as shown in fig...

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  3. A man walks on a straight road form his home to a market 2.5 km away w...

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  4. A particle experiences a constant acceleration for 20 sec after starti...

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  5. If a ball is thrown vertically upwards with speed u, the distance cove...

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  6. A man throws balls with the same speed vertically upwards one after th...

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  7. A stone falls freely such that the distance covered by it in the last ...

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  8. The velocity of a particle is given by v=(5t^(2)-6t+5)m//s where t is ...

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  9. Which of the following equation represents the motion of a body moving...

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  10. The variation of velocity of a particle moving along a straight line i...

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  11. Find the integration of the 1/(2x+1).

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  12. The numerical ratio of displacement to the distance covered is always

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  13. Which of the following velocity-time graphs shows a realistic situatio...

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  14. A particle moves in straight line in same direction for 20 seconds wit...

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  15. A body starts from rest and moves with a uniform acceleration of 10ms^...

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  16. An object travels 10km at a speed of 100 m/s and another 10km at 50 m/...

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  17. A body starting from rest moves along a straight line with a constant ...

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  18. Acceleration-time graph of a body is shown. The corresponding velocity...

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  19. Velocity-time curve for a body projected vertically upwards is with ti...

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  20. A river 4.0 miles wide is following at the rate of 2 miles/hr. The min...

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