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bflsw09 · Sep 16, 2009, 05:28 PM

An arrow is shot into the air at an angle of 60.6 above the horizontal with a speed of 21.4 m/s.

(a) What are the x and y components of the velocity of the arrow 3 s after it leaves the bowstring?

(b) What are the x and y components of the displacement of the arrow during the 3 s interval?

Unknown008 · Sep 17, 2009, 08:09 AM

First, have a sketch. The arrow is shot from the left, and goes to the right.

Find the vertical and horizontal components when the arrow is shot.

Since there is no force acting in the left-right direction, so, the speed along the horizontal (horizontal component) remains the same throughout.

In the up-down direction, you have acceleration due to gravity. You have to use
$v=u+at$ .

Let's take the up direction as positive. The acceleration due to gravity is therefore negative. You have u, the initial vertical velocity. You can find v, the final vertical velocity when time is 3 seconds.

Displacement for the horizontal is given by $s=vt$ where s is the displacement, v the velocity and t the time.

Displacement for the vertical here requires that formula:

$s=ut+\frac12 at^2$

s being the displacement, u, the initial velocity, a the acceleration (don't forget it's negative) and t the time.

I hope this helped! :)

bflsw09 · Sep 17, 2009, 12:19 PM

Got it thanks a lot

Unknown008 · Sep 18, 2009, 09:49 AM

You're welcome! :)