A hot air balloon is ascending vertically at a constant velocity of 5 m s ⁻¹ . A packet is dropped from it when the balloon is at a height of 60 m from the ground. At that exact instant, a person on the ground starts running at a constant speed to catch the packet just before it hits the ground. If the person catches the packet successfully, what was the initial horizontal distance of the person from the drop point?

Step-by-step solution

When the packet is dropped, it inherits the upward velocity of the balloon. Therefore, the initial velocity of the packet is u = +5 m s ⁻¹ (taking upward as positive). The displacement of the packet when it reaches the ground is S = -60 m . Using the second equation of motion for the packet: S = ut - 1 2 gt^2 -60 = 5t - 1 2 (10)t^2 -60 = 5t - 5t^2 t^2 - t - 12 = 0 Solving the quadratic equation: (t - 4)(t + 3) = 0 Since time cannot be negative, t = 4 s . The person runs for this exact duration to catch the packet.