A body is projected at `45^(@)` with a velocity of `20ms^(-1)` has a range of `30m` .The decrease in range due to air resistance is?`(g=10ms^(-2))`

A body projected at 45^(@) with a velocity of "20m/s" has a range of 10m.The decrease in range due to air resistance is g=10ms^(-2)

A body is projected up with a velocity 50 ms^(-1) after one second if acceleration due to gravity disappears then body

A body of mass 4kg is projected vertically up with a velocity of 20ms^(-1) . If the energy spent in overcoming air resistance is 200J ,the maximum height attained by it is? (g=10m/ s^(2) )

A body is projected with a velocity of 30 ms^(-1) at an angle of 30^(@) with the vertical. Find the maximum height, time of flight and the horizontal range.

A body is project at t= 0 with a velocity 10 ms^-1 at an angle of 60 ^(@) with the horizontal .The readius of curvature of its trajectory at t=1s is R. Neglecting air resitance and taking acceleration due to gravity g= 10 ms^-2 , the value of R is :

A body is projected with a velocity 40ms^(-1) at 30^(0) to horizontal.Its initial velocity vector is

A body is projected with a velocity 30 ms^(-1) at an angle of 30^@ with the vertical. Find (i) the maximum height (ii) time of flight and (iii) the horizontal range of the projectile. Take g=10 m//s^@ .

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

A particle of mass 2 kg is projected vertically upward with a velocity of 20 m/s . It attains maximum height of 17 m. The loss in mechanical energy due to air drag is (g = 10 ms^-2 )