A stone is projected from level ground with speed u and ann at angle `theta` with horizontal. Somehow the acceleration due to gravity (g) becomes double (that is 2g) immediately after the stone reaches the maximum height and remains same thereafter. Assume direction of acceleration due to gravity always vertically downwards.
Q. The horizontal range of particle is

A stone is projected from level ground with speed u and ann at angle theta with horizontal. Somehow the acceleration due to gravity (g) becomes double (that is 2g) immediately after the stone reaches the maximum height and remains same thereafter. Assume direction of acceleration due to gravity always vertically downwards. Q. The total time of flight of particle is:

A stone is projected from level ground with speed u and ann at angle theta with horizontal. Somehow the acceleration due to gravity (g) becomes double (that is 2g) immediately after the stone reaches the maximum height and remains same thereafter. Assume direction of acceleration due to gravity always vertically downwards. Q. The total time of flight of particle is:

A stone is projected from level ground with speed u and ann at angle theta with horizontal. Somehow the acceleration due to gravity (g) becomes double (that is 2g) immediately after the stone reaches the maximum height and remains same thereafter. Assume direction of acceleration due to gravity always vertically downwards. Q. The angle phi which the velocity vector of stone makes with horizontal just before hitting the ground is given by:

A stone is projected from level ground with speed u and ann at angle theta with horizontal. Somehow the acceleration due to gravity (g) becomes double (that is 2g) immediately after the stone reaches the maximum height and remains same thereafter. Assume direction of acceleration due to gravity always vertically downwards. Q. The angle phi which the velocity vector of stone makes with horizontal just before hitting the ground is given by:

A body of mass m is projected from ground with speed u at an angle theta with horizontal. The power delivered by gravity to it at half of maximum height from ground is

A particle is projected from ground with speed u and at an angle theta with horizontal. If at maximum height from ground, the speed of particle is 1//2 times of its initial velocity of projection, then find its maximum height attained.

A particle is projected from ground with speed u and at an angle theta with horizontal. If at maximum height from ground, the speed of particle is 1//2 times of its initial velocity of projection, then find its maximum height attained.

A particle of mass m is projected with speed u at angle theta with horizontal from ground. The work done by gravity on it during its upward motion is