Answer:
1.97 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.8 m/s²
Solving the above equation we get
So, the time the package was in the air is 1.97 seconds
Two ships leave a harbor at the same time, traveling on courses that have an angle of 110∘ between them. If the first ship trave
ls at 22 miles per hour and the second ship travels at 34 miles per hour, how far apart are the two ships after 1.2 hours?
1 answer:
Answer:
49.07 miles
Explanation:
Angle between two ships = 110° = θ
First ship speed = 22 mph
Second ship speed = 34 mph
Distance covered by first ship after 1.2 hours = 22×1.2 = 26.4 miles = b
Distance covered by second ship after 1.2 hours = 34×1.2 = 40.8 miles = c
Here the angle between the two sides of a triangle is 110° so from the law of cosines we get
a² = b²+c²-2bc cosθ
⇒a² = 26.4²+40.8²-2×26.4×40.8 cos110
⇒a² = 2408.4
⇒a = 49.07 miles
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