A projectile thrown with velocity v making angle `theta` with vertical gains maximum height H in the time for which the projectile remains in air, the time period is

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

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

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

Establish that projectile thrown making angle theta with horizontal gains a parabolic path.

For a projectile thrown with a velocity v, the horizontal range is (sqrt(3)v^2)/(2g) . The vertical range is v^2/(8g) . The angle which the projectile makes with the horizontal initially is:

For a projectile thrown with a velocity v, the horizontal range is (sqrt(3)v^2)/(2g) . The vertical range is v^2/(8g) . The angle which the projectile makes with the horizontal initially is:

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