A small mailbag is released from a helicopter that is descending steadily at 2.6
ID: 251211 • Letter: A
Question
A small mailbag is released from a helicopter that is descending steadily at 2.61 m/s. After 5.00 s, what is the speed of the mailbag? v = m/s How far is it below the helicopter? d = m What are your answer to parts(a) and (b) if the helicopter is rising steadily at 2.61 m/s? v = m/s d = m A tennis player tosses a tennis bell straight up and then catches it after 2.06s at the same height as the point of release. What is the acceleration of the ball while it is in fight? magnitude m/s^2 direction What is the velocity of the ball when it reaches its maximum height? magnitude m/s direction Find the initial velocity of the ball. m/s upward Find the maximum height it reaches. mExplanation / Answer
5.
(a) Consider the axis (Oy) having the positive sense downward:
v=v0+gt
v0=2.61 m/s
g=9.8 m/s2
v=51.61 m/s
(b) Whithin 5 sec the helicopter descended h(h)=v0*t=13.05 m
The height traveled by the mailbag: h(m)=v0t+gt2/2=135.55 m
The difference is 122.5 m
(c) If consider the axis (Oy) also downward:
v0<0.
The eqs become: v=-v0+gt=49 m/s
h(h)=v0*t=13.05 m (if considering the axis, h(m) is negative, but in this case we are interested in the modulus of the distance it traveled)
h(m)=-v0t+gt2/2=109.45 m
The distance between helicopter and mailbag: h(h)+h(m)=122.5 m
The same result.
6.
(a) g=9.8 m/s2; downward. This means the ball has a negative acceleration while climbing, and a positive acceleration while descending.
(b) v=0 (that's why is the maximum height). Here the sense of velocity changes from upward to downward.
(c) If not friction with the air, time for climbing (tc) = time for descending (td) =t/2=1.03 s
v=v0-gtc/d --> v0=gtc/d~10.1 m/s
(d)
We may use both Galilei formula or the law of movement:
v2=v02-2ghmax --> hmax=v02/2g~5.2 m
hmax=v0tc/d-gtc/d2/2~5.2 m.
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