During takeoff, an airplane climbs with a speed of 195 m/s at an angle of 15° above the horizontal. The speed and angle constitu
te a vector called the velocity. What are the horizontal and vertical components of this velocity?
1 answer:
Answer:
The horizontal component of the velocity is 188 m/s
The vertical component of the velocity is 50 m/s.
Explanation:
Hi there!
Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:
We can find vx using the following trigonometric rule of a right triangle:
cos α = adjacent / hypotenuse
cos 15° = vx / 195 m/s
195 m/s · cos 15° = vx
vx = 188 m/s
The horizontal component of the velocity is 188 m/s
To calculate the y-component we will use the following trigonometric rule:
sin α = opposite / hypotenuse
sin 15° = vy / 195 m/s
195 m/s · sin 15° = vy
vy = 50 m/s
The vertical component of the velocity is 50 m/s.
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Answer:
2. at the lowest point
Explanation:
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E = KE + GPE = const.
Therefore, when KE is maximum, GPE is minimum, and viceversa.
So, the point of the motion where the KE is maximum is where the GPE is minimum: and since the GPE is directly proportional to the heigth of the bob:
we see that GPE is minimum when the bob is at the lowest point,so the correct answer is
2. at the lowest point