Question

An arrow is shot into the air at an angle of 60.0° above the horizontal with a speed of 20.8 m/s.

What is the y-component of the velocity of the arrow 3.0 s after it leaves the bowstring?

Answer 1

Given that launching speed of arrow is:

V0 = 20.8 m/sec at 60 deg above the horizontal

So initial vertical speed of arrow will be, V0y = 20.8*sin 60 deg = 18.0 m/sec

g = acceleration in vertical direction = -9.81 m/sec^2

t = 3.0 sec

Using 1st kinematic equation:

Vy = V0y + a*t

Vy = 18.0 - 9.81*3.0

Vy = -11.43 m/sec = y-component of velocity after 3.0 sec (Here negative sign means velocity is in downward direction)

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