From the top of a tower, a particle is thrown vertically downwards with a velocity of `10 m//s`. The ratio of the distances, covered by it in the `3rd` and `2nd` seconds of the motion is `("Take" g = 10 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))

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))

A ball is thrown vertically downward with a velocity of 20m/s from the top of a tower,It hits the ground after some time with a velocity of 80m/s .The height of the tower is : (g=10m/g^(2))

A particle is projected vertically upwards with an initial velocity of 40 m//s. Find the displacement and distance covered by the particle in 6 s. Take g= 10 m//s^2.

A particle is projected vertically upwards with an initial velocity of 40 m//s. Find the displacement and distance covered by the particle in 6 s. Take g= 10 m//s^2.

A ball is thrown vertically upwards with a speed of 10 m//s from the top of a tower 200 m height and another is thrown vertically downwards with the same speed simultaneously. The time difference between them on reaching the ground is (g=10 m//s^(2))

A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity 20 ms^-1 . The second stone will overtake the first after travelling a distance of (g=10 ms^-2)

A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity 20 ms^-1 . The second stone will overtake the first after travelling a distance of (g=10 ms^-2)