As preparation for this problem, review Conceptual Example 10. The drawing shows
ID: 1521770 • Letter: A
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 123 m/s, and each tank is at the same height of 1.90 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° above the horizontal and the other is flying at an angle of 15.0° below the horizontal. Find the (a) magnitude and (b) direction of the velocity with which the fuel tank hits the ground if it is from plane A. Find the (c) magnitude and (d) direction of the velocity with which the fuel tank hits the ground if it is from plane B. In each part, give the direction as a positive angle with respect to the horizontal.
Explanation / Answer
for plane A :
Consider the motion along X-direction :
Vox = initial velocity in X-direction = 123 Cos15 = 118.8 m/s
a = 0
since acceleration is 0 ,
Vfx = Vox = 118.8 m/s
Consider the motion in Y-direction
Voy = initial velocity in Y-direction = 123 Sin15 = 118.8 m/s
a = - 9.8 m/s2
Y = displacement = -1900 m
Vfy = final velocity
using the equation
Vfy2 = Voy2 + 2 a Y
Vfy2 = (118.8)2 + 2 (-9.8) (-1900)
Vfy = 226.6 m/s
magnitude = sqrt (Vfx2 + Vfy2) = sqrt(118.82 + 226.62) = 255.85 m/s
direction : = tan-1 (Vfy/Vfx) = tan-1(226.6 /118.8) = 62.4
For plane B :
Consider the motion along X-direction :
Vox = initial velocity in X-direction = 123 Cos15 = 118.8 m/s
a = 0
since acceleration is 0 ,
Vfx = Vox = 118.8 m/s
Consider the motion in Y-direction
Voy = initial velocity in Y-direction = -123 Sin15 = -118.8 m/s
a = - 9.8 m/s2
Y = displacement = -1900 m
Vfy = final velocity
using the equation
Vfy2 = Voy2 + 2 a Y
Vfy2 = (-118.8)2 + 2 (-9.8) (-1900)
Vfy = 226.6 m/s
magnitude = sqrt (Vfx2 + Vfy2) = sqrt(118.82 + 226.62) = 255.85 m/s
direction : = tan-1 (Vfy/Vfx) = tan-1(226.6 /118.8) = 62.4
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