**Arrow Hits Apple**

Part A = 6.70 s

Part B = 5.59 s

**Solution Below:**

^{2}for the magnitude of the acceleration due to gravity.

Part A

It’s convenient to just use the formula for the range of a projectile:

R = v^2 * sin(2θ) / g

220 = v^2 * sin(2 * 45) / 9.8

220 = v^2 / 9.8

V^2 = 2,156

v = 46.43 m/s

Now find the x component of the velocity:

vx = sin(θ) * v

vx = cos(45) * 46.43

vx = 32.83 m/s

The time it takes for the arrow to travel 220m is the same as the time it’s in the air:

t = d / v

t = 220 / 32.83

t = 6.70 s

6.70 s

Part B

Start by finding the time it takes for the apple to fall 6.0 m:

d = v_{0} * t + 1/2at^{2}

6 = 0 * t + 1/2 * 9.8 * t^2

6 = 4.9t^2

t^2 = 1.2245

t = 1.1066 s

The arrow needs 6.70 s to reach the tree, so just subtract for the difference:

wait = 6.70 – 1.1066

wait = 5.59 s

So the apple should be dropped 5.59 s after the arrow is shot.

5.59 s