A small block is released from rest from the top of a smooth curved track. The track ends at the ground level. When the block is at a vertical height of 2 m above the ground, its speed is 4 5 m s ⁻¹ . If g = 10 m s ⁻² , the vertical height from which the block was released is:

Step-by-step solution

Let the initial height of the block be H . Applying conservation of mechanical energy between the release point and the point at height h = 2 m : m g H = m g h + 1 2 m v^2 Substituting the given values ( g = 10 m s ⁻² , h = 2 m , v = 4 5 m s ⁻¹ ): 10 H = 10 2 + 1 2 (4 5 )^2 10 H = 20 + 1 2 80 10 H = 20 + 40 = 60 H = 6 m Answer: 6 m