Question

Here we have two planes each about to drop an empty fuel tank. At the moment of release each plane has the same speed of 135 m/s, and each tank is at the same height of 2.00 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 15.0o above the horizontal and the other is flying at an angle of 15.0o below the horizontal. Find the magnitude and direction of the velocity with which the fuel tank hits the ground if it is from (a) plane A and (b) plane B. In each part, give the directional angles with respect to the horizontal.

Answer #1

a) Time of flight for one flying below horizontal can be calculated by using second equation of motion in vertical ,

h = uyt + 0.5gt^2

2000 = 135*sin 15 degree*t + 0.5*9.8*t^2

2000 = 135*0.2588*t + 0.5*9.8*t^2

t = 16.95 s

Final vertical velocity = u_{y} + gt = 135*sin 15 degree
+ 9.8*16.95 = 201 m/s

Horizontal velocity = 135 cos 15 degree = 130.4 m/s

angle at impact = arctan(201/130.4) = 57 degree

magnitude of velocity V = sqrt(201^2+130.4^2) = 239.6 m/s

**Hence magnitude and direction of velocity = 239.6 m/s at
57 degree below horizontal**

b) It will be same for plane A and B as the tank from upward moving plane will have same downward vertical velocity after sometime.

so velocity will be same as **239.6 m/s at 57 degree below
horizontal**

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