A plane is flying horizontally with speed 292 m/s at a height 3880m above the gr

Question

A plane is flying horizontally with speed 292 m/s at a height 3880m above the ground, when a package is dropped from the plane. The acceleration of gravity is 9.8m/s^2 (Neglect air resistance) A second package is thrown downward from the plane with a vertical speed V1 = 58 m/s. What is the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground. -answer in units o m/s
Also, what is the horizontal distance traveled by this second package? -answer in units of m

Explanation / Answer

Let the packet fall 4510 m in t sec =>By s = ut + 1/2gt^2 =>4510 = 0 + 1/2 x 9.8 x t^2 =>t =v920.41 =>t = 30.34 sec =>By R = [Ux] x t =>R = 148 x 30.34 = 4490.06 m 6) By u = v[(Ux)^2 + (Uy)^2] =>u = v[(148)^2 + (63)^2] =>u = 160.85 m/s 7) By s = ut + 1/2gt^2 =>4510 = 63 x t + 1/2 x 9.8 x t^2 =>4.9t^2 + 63t - 4510 = 0 =>t = [-63+/-v{(63)^2 - 4 x 4.9 x (-4510)}]/[2 x 4.9] =>t = [-63 +/- 297.31]/9.8 =>t = 23.91 Thus R = 148 x 23.91 = 3538.63 m

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