An object is projected with a velocity of `20(m)/(s)` making an angle of `45^@` with horizontal. The equation for the trajectory is `h=Ax-Bx^2` where h is height, x is horizontal distance, A and B are constants. The ratio A:B is (g`=ms^-2)`

An object is projected with a velocitiy of 20m//s making an angle of 45^(@) with horizontal. The equation for trajectory is h=Ax-Bx^(2) where h is height x is horizontal distance. A and B are constants. The ratio A.B is x/0.1 . Find value of x.(g=10m//s^(2))

An object is projected with a velocity of 10 m//s at an angle 45^(@) with horizontal. The equation of trajectory followed by the projectile is y = a x - beta x^(2) , the ratio alpha//beta is

A particle is projected with velocity of 10m/s at an angle of 15° with horizontal.The horizontal range will be (g = 10m/s^2)

A particle is projected with velocity 50 m/s at an angle 60^(@) with the horizontal from the ground. The time after which its velocity will make an angle 45^(@) with the horizontal is

A stone is projected with a velocity 20sqrt2 m//s at an angle of 45^(@) to the horizontal.The average velocity of stone during its motion from starting point to its maximum height is

A particle is projected with a velocity of 10sqrt2 m//s at an angle of 45^(@) with the horizontal. Find the interval between the moments when speed is sqrt125 m//s(g=10m//s^(2))

A projectile is thrown with a velocity of 20 m//s, at an angle of 60^(@) with the horizontal. After how much time the velocity vector will make an angle of 45^(@) with the horizontal (in upward direction) is (take g= 10m//s^(2))-

A particle is projected with velocity 20sqrt(2)m//s at 45^(@) with horizontal. After 1s (g=10m//s^(2)) match the following table

A particle is projected with an initial velocity of 200 m//s in a direction making an angle of 30^(@) with the vertical. The horizontal distance covered by the particle in 3s is