An arrow is shot at an angle of θ=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.

1) find ta, the time that the arrow spends in the air.
answer numerically in seconds, to two significant figures.

2) suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree.

Let us solve the Part A first. 

In order to find the time we use the formula for the range of a projectile:

R = v^2 multiply by sin(2θ) divided by g

220 = v^2 multiply by sin(2 multiply by 45) divided by 9.8

220 = v^2 divided by 9.8

V^2 = 2,156

v = 46.43 m/s

Now we will find the x component of the velocity:

vx = sin(θ) multiply by v

vx = cos(45) multiply by 46.43

vx = 32.83 m/s

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

time = d divided by  v

time = 220 divided by 32.83

time = 6.70 seconds

6.70 s.

Now we will solve the part B.

We will start by finding the time it takes for the apple to fall 6.0 m

d = v0 multiply by t + 1 divided by 2 at 2 square

6 = 0 multiply by t + 1 divided by 2 multiply by 9.8 multiply by t^2

6 = 4.9 multiply by t^2

t^2 = 1.2245

t = 1.1066 s

Thus the arrow needs 6.70 s to reach the tree, so we will just subtract for the difference:

wait = 6.70 – 1.1066

wait = 5.59 s

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

I hope it helped

35 is an outlier

step-by-step explanation: