A hot air balloon is rising vertically upwards at a constant uniform speed. An observer is standing on the horizontal ground at a fixed distance from the launch pad. Initially, the angle of elevation of the balloon as seen by the observer is 30^ . After 10 minutes, the angle of elevation becomes 45^ . The further time taken (in minutes) for the angle of elevation to become 60^ is :

Step-by-step solution

Let the distance of the observer from the launch pad be d and the uniform speed of the balloon be v . Let the heights of the balloon at the three instants be h₁ , h₂ , and h₃ . From the given angles of elevation: h₁ = d 30^ = d 3 h₂ = d 45^ = d h₃ = d 60^ = d 3 The distance covered by the balloon in the first 10 minutes is: h₂ - h₁ = d - d 3 = d ( 3 -1 3 ) Since the speed is uniform, v = h₂ - h₁ 10 = d( 3 -1) 10 3 . The further distance to be covered for the angle of elevation to become 60^ is: h₃ - h₂ = d 3 - d =