A projectile has the maximum range 500 m . If the projectile is thrown up an inclined plane of 30^ with the same (magnitude) velocity, the distance covered by it along the inclined plane will be

Step-by-step solution

Since, range is maximum, therefore =45^ . Hence, using R=u^2 2 / g . We find 500= u^2 g , i.e. u=[500 g]^ 1 2 . Distance covered along the inclined plane can be obtained using relation v^2-v₀^2=2 a x 0-u^2=2 (-g 30^ ) x [ a=-g 30^ ] x= u^2 g =500 ~m