A body is projected vertically up with a speed of `20ms^(-1)` .The distance travelled by it in "3" seconds it `(g=10ms^(-2))`

A ball is projected vertically up with speed 20 m/s. Take g=10m//s^(2)

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]:-

A body is projected horizontally with speed 20 ms^(-1) . The speed of the body after 5s is nearly

A body of mass 4kg is projected vertically up with a velocity of 20ms^(-1) . If the energy spent in overcoming air resistance is 200J ,the maximum height attained by it is? (g=10m/ s^(2) )

A man standing on the edge of the roof of a 20 m tall building projects a ball of mass 100 gm vertically up with a speed of 10 ms^(-1) . The kinetic energy of the ball when it reaches the ground will be [g = 10 ms^(-2)]

A body is projected at an angle of 30^(@) with the horizontal and with a speed of 30ms^(-1) . What is the angle with the horizontal after 1.5 second? (g=10ms^(-2))

[" A body is thrown up with "],[" velocity of "24.8ms^(-1)" .Distance "],[" travelled by the body in its last "],[" second of upward journey is "],[g=10ms^(-2)" ) "]