Robin Hood is walking through Sherwood Forest when he comes upon Sheriff of Nottingham. The Sheriff tells Robin Hood that he is going to arrest hum for numerous misdeeds. Robin pulls out his bow, loads an arrow, puls the bow back 0.70 m, and shoots an arrow into the Sheriff's hat, knocking the hat with the arrow lodged in it off his head. The hat's mass is 0.5 kg. It takes 4 N to pull Robin's bow back 0.10 m, and the arrow has a mass of 0.20 kg. (A) What is the speed of the hat with the arrow in it just fter it is struck? Assume the arrow travels horizontally between Robin and the Sheriff and neglect air resistance. (B) If the Sheriff is 2.0 m high, how far behind him will his hat with the arrow in it hit the ground?
I found the answer to (A) to be 2.8 m/s how do I get (B)?
Yes, I get 2.83 m/s for A
Object of mass m (which happens to be .7 kg but it will not matter) is travelling horizontal at 2.83 m/s NO vertical speed yet but 2 m high.
How long will it take to fall to the ground?
d = 2 = (1/2) (9.8) t^2
t^2 = .408
t = .639 seconds to landing
How far did it go horizontal at 2.83 m/s for .639 s?
distance horizontal = 1.80 m
how did you get 2.8 m/s for A?
To find the distance at which the hat with the arrow in it will hit the ground behind the Sheriff, we need to use the principles of projectile motion.
First, let's analyze the motion of the arrow when it hits the hat. We know the initial speed of the arrow (which is equal to the final speed of the hat) is 2.8 m/s. We also know that the hat and arrow move horizontally without any vertical acceleration due to neglecting air resistance. Therefore, the vertical component of the hat's motion only involves free fall under gravity, which will cause it to fall vertically downward.
Now, let's find the time it takes for the hat with the arrow in it to hit the ground. We can use the equation of motion for vertical free fall:
h = (1/2)gt^2
Here, h is the height of the hat, which is 2.0 m, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight.
Rearranging the equation, we get:
t = sqrt(2h / g)
Substituting the values, we have:
t = sqrt(2 * 2.0 / 9.8) = sqrt(0.4082) ≈ 0.638 s
Now that we have the time of flight, we can find the horizontal distance the hat with the arrow will travel during that time. Since there is no horizontal acceleration, the distance is simply equal to the horizontal component of the initial velocity multiplied by the time:
d = v * t
Substituting the values, we have:
d = 2.8 * 0.638 ≈ 1.78 m
Therefore, the hat with the arrow will hit the ground approximately 1.78 meters behind the Sheriff.
Note: It's important to mention that this calculation neglects any potential vertical motion of the hat after it is hit by the arrow.