A helicopter is flying in a straight line over a level field at a constant speed
ID: 1589536 • Letter: A
Question
A helicopter is flying in a straight line over a level field at a constant speed of 5.8 m/s and at a constant altitude of 9.1 m. A package is ejected horizontally from the helicopter with an initial velocity of 15.0 m/s relative to the helicopter, and in a direction opposite the helicopter's motion.
Find the initial velocity of the package relative to the ground. -9.20 m/s You are correct.
What is the horizontal distance between the helicopter and the package at the instant the package strikes the ground? Incorrect. 7/8 Previous Tries
What angle does the velocity vector of the package make with the ground at the instant before impact, as seen from the ground? 5.54×10^1 deg You are correct.
1 Incorrect. (Try 1) 4.0m 2 Incorrect. (Try 2) 6.28m 3 Incorrect. (Try 3) 10.24m 4 Incorrect. (Try 4) 9.37m 5 Incorrect. (Try 5) 2.7m 6 Incorrect. (Try 6) 27.86m 7 Incorrect. (Try 7) 38.63mExplanation / Answer
initial Velocity of helicopter ,Vh= 5.8 m/s
height,h = 9.1 m
relative velocity of package w.r.t to helicopter ,Vph = 15 m/s
i) Vph = Vp+Vh ( because both are in opposite direction)
15 = Vp + 5.8
Vp ( initial velocity of package) = 9.2 m/s
Vpg = relative velocity ofpackge w.r.t ground
Vpg = Vp - Vg
= 9.2 - 0 = 9.2 m/s ( Vg = velocity of ground = 0 m/s at rest)
ii) time taken to hit the ground,t :
by taking vertical motion, Vy2 - uy2 = 2*a*h ( uy = 0 initial velocity in y-direction = 0)
Vy2 - 0 = 2*9.8*9.1
Vy = 13.35 m/s
V = u+a*t
13.35 = 0 + 9.8*t
t = 1.36 sec
horizontal distance travelled by package = 9.2*1.36 = 12.512 m ( dis = vel*t)
horizontal distance travelled by helicopter = 5.8*1.36 = 7.888 m
distance between them = 12.512 + 7.888 = 20.4 m
angle , tanB = V_y / u _x
= 13.35 / 9.2
B = 55.42 degree
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