A body is thrown vertically upward with a velocity of 9.8 m/s. Calculate the maximum height attained by the body.(g=9.8 m/s^2)

A ball is thrown vertically upwards with a velocity of 49 m//s . Calulate (i) The maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth.

A stone is thrown vertically upwards with initial velocity of 14 m s^-1 . The maximum height it will reach is [g= 9.8 m s^-2]

A stone is thrown vertically upwards with initial velocity of 14 m s^-1 . The maximum height it will reach is [g = 9.8 m s^-2]

A stone is thorwn vertically upwards with initial velocity of 40m//s . Taking g=10m//s^2 find the maximum height and total distance covered by stone

A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is

A body is thrown with velocity of 49 m s^-1 at an angle of 30^@ with the horizontal, what is maximum height attained by body? , A. 20.625 B. 61.25 C. 30.625 D. 41.25

A stone is thrown vertically upward with a velocity of 30 m//s . How high will it rise? After how much time will it return to ground ? [Take g = 10 m//s^2]

A body projected vertically upwards with a velocity of u returns to the starting point in 6 second, If g= 9,8 m s^-2 the value of 'u' is

An object is thrown vertically upwards and rises to a height of 10 m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.

A body is projected vertically upwards with a velocity of 10 ms^-1 and another body is projected simultaneously from the same point with a velocity of 20 m s^-1 at an angle of pi/6 with the horizontal. The distance between the two bodies after one second from the time of projection is (Acceleration due to gravity is 10 ms^-2 )