An observer on the ground is x meters from the base of the launch pad of a rocket, which is at the same level as the observer. A
few seconds after the rocket takes off vertically, the observer sees its tip at an angle of q° from the horizontal. How far above the ground is the tip of the rocket at that instant? Assume that the ground is level.
x tan q
x sin q
x cos q
x tan q
x sin q
x cos q
2 answers:
Answer:
Tip of the rocket is x tan q meters above the ground.
Step-by-step explanation:
From the figure attached, at any moment after the launch of the rocket, tip of the rocket is at A.
AB is the height of the rocket tip from the ground and observer is x meters away from the launch pad.
Then tan q =
tan q =
AB = x tan q meters.
Tip of the rocket is x tan q meters above the ground.
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Answer:
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Step-by-step explanation:
Let's assume that
- the velocity of first car,
- and the velocity of second car
=> speed of first car,
= 32.01 units
and speed of second car,
= 30 units
= 2.01 units
Hence, 2nd car is running faster than the first car by 2.01 units.
When
When
Adding the two equations together gives
Now, you have
By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that
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