An arrow is shot into the air at an angle of 59.6° above the horizontal with a speed of 20.7 m/s. What is the x-component of the velocity of the arrow 3.20 s after it leaves the bowstring?
Question
An arrow is shot into the air at an angle of 59.6° above the horizontal with a speed of 20.7 m/s. What is the x-component of the velocity of the arrow 3.20 s after it leaves the bowstring?
Solution
The x-component of the velocity of the arrow can be calculated using the initial velocity, the angle of projection, and the cosine function.
The x-component (Vx) of the velocity is given by the equation:
Vx = V * cos(θ)
where: V is the initial velocity (20.7 m/s), and θ is the angle of projection (59.6°).
First, convert the angle from degrees to radians because the trigonometric functions in most calculators use radians.
1 radian = 180/π degrees So, 59.6° = 59.6 * π/180 radians = 1.04 radians
Then, substitute the values into the equation:
Vx = 20.7 m/s * cos(1.04 rad)
The x-component of the velocity does not change over time in ideal projectile motion (no air resistance), so the x-component of the velocity 3.20 s after the arrow leaves the bowstring is the same as the initial x-component of the velocity.
Therefore, calculate the value to find the x-component of the velocity.
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