An archer stands a horizontal distance d= 55 m away from a tree sees an apple hanging from the tree at h= 12 m above the ground. The archer chooses an arrow and prepares to shoot. The arrow is initially 1.5 m above the ground. Just as the archer shoots the arrow with a speed of 80 m/s , the apple breaks off and falls straight down. A person of height 2.0 m is standing directly underneath the apple. The arrow pierced the apple. Ignore air resistance, and use g=9.81 m/s2 for the acceleration of gravity.
(a) What angle did the archer aim the arrow? enter your answer in degrees
(b) How high above the person's head did the arrow hit the apple?
theta =12 y=7.97
y is correct but theta is wrong
To solve this problem, we need to break it down into two parts:
Part 1: Determine the time it takes for the arrow to reach the height of the apple.
We can use the equation of motion for vertical motion to find the time it takes for the arrow to reach the height of the apple:
h = h0 + v0t + (1/2)at^2
Here:
h = height of the apple above the ground = 12 m
h0 = initial height of the arrow above the ground = 1.5 m
v0 = initial vertical velocity of the arrow = 0 m/s (since the arrow is initially at rest vertically)
a = acceleration due to gravity = -9.81 m/s^2 (negative because it acts downward)
t = time taken to reach the height of the apple
Rearranging the equation, we get:
t = sqrt(2(h - h0) / a)
Substituting the given values, we have:
t = sqrt(2(12 - 1.5) / -9.81)
Calculating this gives us t ≈ 1.29 seconds.
Part 2: Determine the angle at which the archer aimed the arrow.
To find the angle at which the archer aimed the arrow, we can use the equation of motion for horizontal motion:
d = v0x * t
Here:
d = horizontal distance to the tree = 55 m
v0x = initial horizontal velocity of the arrow
t = time taken to reach the height of the apple (which we found to be 1.29 seconds)
Since the horizontal velocity of the arrow remains constant throughout its motion, we have:
v0x = d / t
Substituting the given values, we have:
v0x = 55 / 1.29
Calculating this gives us v0x ≈ 42.64 m/s.
Now, we can find the angle at which the archer aimed the arrow using trigonometry. The horizontal velocity and the initial vertical velocity form the legs of a right triangle, with the resultant velocity (the speed of the arrow, given as 80 m/s) as the hypotenuse.
tanθ = v0x / v0
Substituting the given values, we have:
tanθ = 42.64 / 80
Taking the inverse tangent (arctan) of both sides, we can find the angle:
θ = arctan(42.64 / 80)
Calculating this gives us θ ≈ 27.55 degrees.
Therefore, the archer aimed the arrow at an angle of approximately 27.55 degrees.
Part 3: Determine how high above the person's head the arrow hit the apple.
Since the arrow pierced the apple, we can assume that the arrow's height is the same as the apple's height when they intersect.
Using the equation of motion for vertical motion, we can find the height above the person's head where the arrow hit the apple:
h = h0 + v0yt + (1/2)at^2
Here:
h = final height of the arrow above the ground (also her height above the person's head)
h0 = initial height of the arrow above the ground = 1.5 m
v0y = initial vertical velocity of the arrow
a = acceleration due to gravity = -9.81 m/s^2 (negative because it acts downward)
t = time taken to reach the height of the apple (which we found to be 1.29 seconds)
Since the arrow was shot vertically upwards, its initial vertical velocity is positive:
v0y = 80 m/s (as given)
Substituting the given values, we have:
h = 1.5 + (80 * 1.29) + (1/2)(-9.81)(1.29)^2
Calculating this gives us h ≈ 65.82 m.
Therefore, the arrow hit the apple at a height of approximately 65.82 meters above the person's head.