A rescue plane flying horizontally drops a supply packet. The packet hits the ground exactly 10 s after it is released. If the straight-line distance from the point of release to the point of impact is 1300 m , what was the initial speed of the plane? (Take g = 10 m/s ^2 and neglect air resistance)

Step-by-step solution

Let the initial speed of the plane be u . The horizontal displacement after time t is x = ut . The vertical displacement of the packet in time t = 10 s is: y = 1 2 gt^2 = 1 2 10 (10)^2 = 500 m The total straight-line displacement D is given by D = x^2 + y^2 . Given D = 1300 m , we can find the horizontal displacement x : 1300 = x^2 + 500^2 x^2 = 1300^2 - 500^2 = 1690000 - 250000 = 1440000 x = 1200 m Since x = ut , the initial speed of the plane is: u = x t = 1200 10 = 120 m/s Answer: 120 m/s