A bus starting from rest moves with a uniform acceleration for 2 minutes. If the velocity at the end of 2 minutes is 12 m/s, its
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<h2>Answer:</h2>
<em><u>Velocity of throwing arrow = 43.13 m/s.</u></em>
<h2>Explanation:</h2>In the question,
Let us say the height from which the arrow was shot = h
Distance traveled by the arrow in horizontal = 61 m
Angle made by the arrow with the ground = 2°
So,
From the <u>equations of the motion</u>,
Now,
Also,
Finally, the angle made is 2 degrees with the horizontal.
So,
Final horizontal velocity = v.cos20°
Final vertical velocity = v.sin20°
Now,
u = v.cos20° (No acceleration in horizontal)
Also,
So,
We can say that,
<em><u>Therefore, the velocity with which the arrow was shot by the archer is 43.13 m/s.</u></em>
Answer:
The corridor's distance is "90 m".
Explanation:
- She heads in the east directions but creates the first pause, meaning she crosses the distance 'x' in step 1.
- Now, provided that perhaps the distance by her to another fountain or waterfall just after the first stop is twice as far away she traveled.
- Because she moved the distance of 'x,' then, therefore, her distance towards the fountain of '2x.' She casually strolls and once again pauses 60 m beyond her stop.
- The gap about her to the waterfall during that time approximately twice the distance and her to the eastern end of the hallway.
- Assume her gap from either the east end of the platform seems to be 'y' at either the second stop, after which '2y' may become the distance between the 2nd pause and the waterfall.
Now,
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The total distance of the corridor will be:
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