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 particle is projected at an angle with horizontal. If T is the time of flight and R is a horizontal range, then theta is

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

A ball is thrown such that the time of flight is 5 s and the horizontal range is 200 m . Find the magnitude and direction of velocity of projection.

A stone is thrown with a velocity v at an angle theta with the horizontal.Its speed when it makes an angle beta with the horizontal is

A javelin thrown into air at an angle with the horizontal has a range of 200 m. If the time of flight is 5 second, then the horizontal component of velocity of the projectile at the highest point of the trajectory is

a ball is thrown with 25 m//s at an angle 53^(@) above the horizontal. Find its time to fight, maximum height and range.

A ball is projected from the ground at angle theta with the horizontal. After 1 s , it is moving at angle 45^@ with the horizontal and after 2 s it is moving horizontally. What is the velocity of projection of the ball ?

A projectile is thrown with velocity u making an angle theta with the horizontal. Its time of flight on the horizontal ground is 4 second. The projectile is moving at angle of 45^(@) with the horizontal just one second after the projection. Hence the angle theta is (take g=10m//s^(2) )

A particle is projected at an angle theta =30^@ with the horizontal, with a velocity of 10ms^(-1) . Then