If direction of velocity of a projectile makes an angle of `37^(@)` with the horizontal,at a time which is `1s` before the time of reaching the maximum height,then velocity at highest point,will be `(g=10m/s^(2))`

A body is thrown with a velocity of 9.8 m/s making an angle of 30^(@) with the horizontal. It will hit the ground after a time

A bomb is fired from a cannon with a velocity of 1000 ms^(-1) making an angle of 30^(@) with the horizontal (g = 9.8 m//s^(2)). What is the time taken by the bomb to reach the highest point ?

A particle of mass 1 kg is projected at an angle of 30^(@) with horizontal with velocity v = 40 m/s . The change in linear momentum of the particle after time t = 1 s will be (g = 10 m// s^(2) )

A projectile is thrown at an angle of 30^(@) with a velocity of 10m/s. the change in velocity during the time interval in which it reaches the highest point is

Velocity of a stone projected, 2 second bofore it reaches the maximum height makes angle 53^@ with the horizontal then the velocity at highest point will be

Velocity of a stone projected, 2 second bofore it reaches the maximum height makes angle 53^@ with the horizontal then the velocity at highest point will be

A projectile is projected from the ground by making an angle of 60^@ with the horizontal. After 1 s projectile makes an angle of 30^@ with the horizontal . The maximum height attained by the projectile is (Take g=10 ms^-2)

A projectile is thrown with velocity v at an angle theta with the horizontal. When the projectile is at a height equal to half of the maximum height,. The vertical component of the velocity of projectile is.

A particle of mass m is projected at an angle alpha to the horizontal with an initial velocity u . The work done by gravity during the time it reaches its highest point is

For a projectile, the ratio of maximum height reached to the square of flight time is (g = 10 ms^(-2))