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 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 ball projected with a velocity of 28m/sec has a horizontal range 40m. Find the two angles of projection.

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 )

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 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 projectile thrown with an initial velocity of 10 ms^(-1) at an angle with the horizontal, has a range of 5 m. Taking g=10 ms^(-2) and neglecting air resistance, what will be the estimated value of alpha ?

A body is projected with an initial velocity 20 m//s at 60^(@) to the horizontal. Its initial velocity vector is "___" (g=10 m//s^(2))

At what angle should a body he projected with a velocity 24 ms^(-1) just to pass over the obstacle 14 m high at a distance of 24 m [Take g = 10ms^(-2) ]

A particle is projected at an angle of 45^(@) with a velocity of 9.8 ms^(-1) . The horizontal range will be (Take, g = 9.8 ms^(-2))