A stone is thrown at an angle `theta` to the horizontal reaches a maximum height H. Then the time of flight of stone will be:

A projectile is thrown at an angle of theta=45^(@) to the horizontal, reaches maximum reaches a maxium height of 16m , then

A projectile is thrown at angle beta with vertical.It reaches a maximum height H .The time taken to reach the hightest point of its path is

A stone projected with a velocity u at an angle q with the horizontal reaches maximum heights H_(1) . When it is projected with velocity u at an angle (pi/2-theta) with the horizontal, it reaches maximum height H_(2) . The relations between the horizontal range R of the projectile, H_(1) and H_(2) , is

A ball of mass m is thrown vertically upwards. Another ball of same mass is thrown at an angle theta to the horizontal. If the time if flights for both is same, the ratio of maximum height attained by them

A particle is projected from the ground with velocity u at angle theta with horizontal.The horizontal range,maximum height and time of flight are R,H and T respectively.Now keeping u as fixed, theta is varied from 30^(@) to 60^(@) .Then

A stone is thrown vertically up from the ground. It reaches a maximum height of 50 m in 10 sec. After what time it will reach the ground from maximum height position ?

A stone projected from ground with certain speed at an angle theta with horizontal attains maximum height h_(1) when it is projected with same speed at an angle theta with vertical attains height h_(2) . The horizontal range of projectile is

A stone is projected at an angle alpha to the horizontal from the top of a tower of height 3h If the stone reaches a maximum height h above the tower, then calculate the distance from the foot of the tower where particle strikes the foot of the where particle strikes the ground.