A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?
Answers
Answered by
0
Answer:
since the package is dropped
therefore initial velocity u=0
height =80m
3rd eq of kinematics
v²=u²+2gh
v²=0+ 80×2×10
v²= 1600 v=40m/s
1st eq of kinematics for time
v=u+gt
40=0×10t
therefore t= 4s
speed = 40m/s
time= 4s
(you can take g as 9.8 aswell)
Answered by
0
Explanation:
verified
Verified by Toppr
Given, for the package:
v
0
=12m/s
h=80m=y
0
Then,
y−y0=v
0
t−
2
1
gt
2
⇒0−80m=(12m/s)t−
2
1
(9.8m/s
2
)t
2
⇒(4.9m/s
2
)t
2
−(12m/s)t−80m=0
⇒(4.9m/s
2
)t
2
−(12m/s)t−80m=0
⇒t=5.45s, or t=−3.00s
∴t=5.45s
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