A projectile is projected with the initial velocity `(6i + 8j) m//s`. The horizontal range is `(g = 10 m//s^(2))`

A body is projected with a initial velocity of (8hati+6hatj) m s^(-1) . The horizontal range is (g = 10 m s^(-2))

A body is projected with a initial velocity of (8hati+6hatj) m s^(-1) . The horizontal range is (g = 10 m s^(-2))

What are the two angles of projection of a projectile projected with velocity of 30 m//s , so that the horizontal range is 45 m . Take, g= 10 m//s^(2) .

What are the angles of projection of a projectile projected with velocity 30 ms^(-1) , so that the horizontal range is 45 m ? Take g = 10 ms^(-2) .

What are two angles of projection of a projectile projected with velocity 30 m/s , so that the horizontal range is 45m . Take g=10 m/s^2 .

A projectile is projected from top of a wall 20m height with a speed of 10m/s horizontally. Find range (g = 10m/s^(2))

What are the two angles of projection of a projectile with velocity 30 m s^-1 , so that the horizontal range is 45 m.Take g=10 m s^-2 .

A projectile is projected with initial velocity (6hati+8hatj) m/s.Minimum time after which its velocity becomes parallel to the line 4y-3x+2=0, is ?( bar(g)=-10hatj m/ s^(2) )