The velocity-time ( v-t ) graph of a particle projected vertically downwards from the top of a tower is a straight line starting from v = -20 m s ⁻¹ at t = 0 and ending at v = -50 m s ⁻¹ at t = 3 s . At t = 3 s , the particle hits the ground and its velocity instantaneously becomes zero. The height of the tower is : (Take g = 10 m s ⁻² )

Step-by-step solution

The displacement of the particle is given by the area under the velocity-time graph. The graph forms a trapezium with parallel sides of magnitudes 20 m s ⁻¹ and 50 m s ⁻¹ , and a height (time interval) of 3 s . Magnitude of displacement = 1 2 ( sum of parallel sides ) time s = 1 2 (20 + 50) 3 s = 1 2 70 3 = 105 m Alternatively, using the equation of motion s = ut + 1 2 at² with u = 20 m s ⁻¹ (downwards), a = 10 m s ⁻² (downwards), and t = 3 s : s = 20(3) + 1 2 (10)(3)² = 60 + 45 = 105 m . Thus, the height of the to