A projectile is thrown with velocity v making an angle `theta` with the horizontal. It just crosses the top of two poles,each of height h, after 1 seconds 3 second respectively. The time of flight of the projectile is

A particle is thrown with velocity 9u) making an angle theta with the vertical. It just crosses the top of two poles each height (h) after 1 s and 3 s respectively . Fing the maximum hight of projecile. G= 9.8 ms//s^2 .

A projectile is fired with a velocity V making an angle theta with the horizontal. Show that its trajectory is parabolic.

A projectile is projected with velocity u making an angle theta with the horizontal, the equation of the path of the projectile is given by

A projectile is projected with velocity u making angle theta with horizontal direction, find : time of flight.

A projectile is thrown with speed u making angle theta with horizontal at t=0 . It just crosses the two points at equal height at time t=1 s and t=3 sec respectively. Calculate maximum height attained by it. (g=10 m//s^(2))

A projectile is thrown with speed u making angle theta with horizontal at t=0 . It just crosses the two points at equal height at time t=1 s and t=3 sec respectively. Calculate maximum height attained by it. (g=10 m//s^(2))

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 projectile is thrown with speed u making angle theta with horizontal at t =0 . It just crosses two points of equal height at time t = 1 s and t = 3 s respectively. Calculate the maximum height attained by it ? (g = 10 m//s^(2) )