<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>
The answer to your question is A
Answer:

Explanation:
Given that,
Initial angular velocity, 
Acceleration of the wheel, 
Rotation, 
Let t is the time. Using second equation of kinematics can be calculated using time.

Let
is the final angular velocity and a is the radial component of acceleration.

Radial component of acceleration,

So, the required acceleration on the edge of the wheel is
.
Answer:
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Explanation:
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