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
a. The time should be 6.70 seconds.
b. The time should be 5.50 seconds.
a.
[tex]R = v^2 \times sin(2\theta ) \div g\\\\220 = v^2 \times sin(2 \times 45)\div 9.8\\\\220 = v^2 \div 9.8[/tex]
[tex]V^2 = 2,156[/tex]
v = 46.43 m/s
Now the x component of the velocity is
[tex]vx = sin(\theta) \times v\\\\vx = cos(45) \times 46.43[/tex]
vx = 32.83 m/s
Now finally the time is
time = [tex]220 \div 32.83[/tex]
time = 6.70 seconds
b.
[tex]d = v0 \times t + 1 \div 2 at 2^2\\\6 = 0 \times t + 1 \div 2 \times 9.8 \times t^2\\\\6 = 4.9\times t^2[/tex]
[tex]t^2 = 1.2245[/tex]
t = 1.1066 s
Now
wait = 6.70 – 1.1066
wait = 5.59 s
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