A body projected with velocity 30 `ms^(-1)` reaches its maximum height in 1.5 s. Its range is (g=`10ms^(-2)`)

A body is projected with a velocity vecv =(3hati +4hatj) ms^(-1) The maximum height attained by the body is: (g=10 ms^(-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) .

A ball is kiched at an angle 30^(@) with the verical. If the horizontal componet of its velocity is 20 ms^(-1) , find the maximum hight and hrizontal range. Use = 10 ms^(-2) .

A body is projected with velocity u so that its horizontal range is twice the maximum height.The range is

A body is thrown with velocity (4i+3j) metre per second Its maximum height is (g=10ms^(-2))

A body is projected from the ground with a velocity v = (3hati +10 hatj)ms^(-1) . The maximum height attained and the range of the body respectively are (given g = 10 ms^(-2))

A body is projected with a velocity 60 ms ^(-1) at 30^(@) to horizontal . Its initial velocity vector is

A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is 10 ms^-1 , then the maximum height attained by the stone is ( g= 10 ms^(-2) )

At what angle of elevation , should a projectile be projected with velocity with velocity 20 ms^(-1) , so as to reach a maximum height of 10 m ?

A body is projected with a velocity of 20 ms^(-1) in a direction making an angle of 60^(@) with the horizontal. Determine its (i) position after 0.5 s and (ii) the velocity after 0.5s .