A ball is thrown at an angle `theta` with the horizontal and the range is maximum. The value of `tantheta` is:-

A projectile is thrown at an angle theta with the horizontal and its range is R_(1) . It is then thrown at an angle theta with vertical anf the range is R_(2) , then

A player kicks up a ball at an angle theta to the horizontal. The horizontal range is maximum when theta equals-

A ball is thrown at an angle theta with the horizontal. Its time of flight is 2 s. Its horizontal range is 100 m. What is the horizontal component of velocity of projection ? Take g =10 ms^(-2) .

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:

The maximum height attained by a projectile when thrown at an angle theta with the horizontal is found to be half the horizontal range. Then theta is equal to

The maximum height attained by a projectile when thrown at an angle theta with the horizontal is found to be half the horizontal range. Then theta is equal to

A ball is thrown at angle theta and another ball is thrown at an angle (90^(@)-theta) with the horizontal direction from the same point with velocity 39.2 ms^(-1) . The second ball reaches 50 m higher than the first ball. Find their individual heights. Take g = 9.8 ms^(-2) .