1.An arrow is shot at an angle of ?=45? above the horizontal. The arrow hits a tree a horizontal distance D=220m away, at the same height above the ground as it was shot. Use g=9.8m/s2 for the magnitude of the acceleration due to gravity.

A) Find ta, the time that the arrow spends in the air.

It’s convenient to 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

The answer is 6.70 s.

B) Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree?

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

d = v0 * t + 1/2at2
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 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.

The answer is 5.59 s.

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