An arrow is shot into the air at an angle of 61.8° above the horizontal with a speed of 21.6 m/s.
(a) What are the x- and y-components of the velocity of the arrow 3.1 s after it leaves the bowstring?
(b) What are the x- and y-components of the displacement of the arrow during the 3.1 s interval?
To solve this problem, we need to break down the motion of the arrow into horizontal and vertical components. We can use the equations of motion to find the x- and y-components of the velocity and displacement.
(a) To find the x- and y-components of the velocity at any instant, we can use the following equations:
Vx = V * cos(θ)
Vy = V * sin(θ)
where Vx is the horizontal component of the velocity, Vy is the vertical component of the velocity, V is the magnitude of the velocity (21.6 m/s), and θ is the angle of launch (61.8°).
Substituting the given values into the equations:
Vx = 21.6 m/s * cos(61.8°)
Vy = 21.6 m/s * sin(61.8°)
Calculating these values will give us the x- and y-components of the velocity.
(b) To find the x- and y-components of the displacement during the 3.1 s interval, we can use the equations of motion:
Sx = Vx * t
Sy = (Vy * t) + (0.5 * g * t^2)
where Sx is the horizontal displacement, Sy is the vertical displacement, Vx and Vy are the x- and y-components of the velocity respectively, t is the time interval (3.1 s), and g is the acceleration due to gravity (9.8 m/s^2).
Substituting the given values into the equations:
Sx = (V * cos(θ)) * t
Sy = (V * sin(θ) * t) - (0.5 * g * t^2)
Calculating these values will give us the x- and y-components of the displacement during the 3.1 s interval.
Note that we are assuming no air resistance in this problem.