As preparation for this problem, review Conceptual Example 10. The drawing shows

Question

As preparation for this problem, review Conceptual Example 10. The drawing shows two planes each dropping an empty fuel tank. At the moment of release each plane has the same speed of 127 m/s, and each tank is at the same height of 4.77 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15.0

Explanation / Answer

The horizontal velocity of the planes is 135*cos(15) = 130,3 m/sec The vertical velocity is 135*sin(15) = 34,9 m/sec up, respectively down. 1) the one tank that goes up with 34,9 m/sec reaches a height of v^2/2g = 62 m from the point of release. So then it drops down 1062 m to the ground. The vertical velocity it reaches is v = v(2s*g) = v(2*1062*9,81) = 144,34 m/sec. 2) the other tank from the plane going down drops from 1000 m and reaches a vertical speed of v = 140,07 m/sec plus the speed downwards given to it by the plane = 140,07 + 34,9 = 174,97 m/sec. The angle alpha with the ground when hitting the ground is tan(alpha) = vertical velocity/horizontal velocity = 144,34/130,3 --> alpha = 47,99° for the tank from the plane going up, respectively = 174,97/130,3 ---> alpha = 53,32° for the tank from the plane going down. The absolute values of velocity are derived by Pythagoras from vertical and horizontal speeds= 194 m/sec for the tank from the up plane and 218 m/sec for the tank from the down plane. Formulas: for the accelerated vertical movement: s = 1/2 gt^2 and v = g*t--> t = v/g s = 1/2 g(v^2/g^2) = 1/2 v^2/g v^2 = 2sg v = v(2sg)

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