An archer shoots an arrow with a velocity of 45 m/s at an angle of 50 degrees with the horizontal.An assistant standing on the level ground 150m downrange from the launch point throws an apple straight up with the minimum intial speed necessary to meet the path of the arrow. (a) What is the initial speed of the apple? (b) At what time after the arrow launch should the apple be thrown so that the arrow hits the apple?
The total range of the arrow is
L=vₒ²•sin2α/g =45²•sin100°/9.8 =203.5 m
R=150 m > L/2=101.75 m
=> the arrow is at its descending path
The time of the motion to the point of 'arrow-apple' meeting is
t₁=R/v(x)=R/v₀(x)= R/ v₀•cos α =150/45•cos50°=5.2 s
The time of the motion to the highest point is
t₀= vₒ•sinα/g=45•sin50°/9.8 =3.5 s
The time of the motion between the highest point and the point of meeting is
t= t₁ - t₀ =5.2 -3.5 = 1.7 s.
The maximum height of the arrow is
H= vₒ²•sin²α/2g =
= 45²•sin²50°/2•9.8=60.6 m/
The vertical downward displacement is
Δh=gt²/2=9.8•1.7²/2 =14.2 m
The position of the meeting point is
H =60.6-14.2 = 46.4 m (above the ground level)
Min. initial speed will be if the apple and the arrow meet at the apple’s top point
H=V₀t₂-gt₂²/2,
0= V₀-gt₂,
H=V₀²/2g =>
V₀=sqrt(2gH) = sqrt(2•9.8•46.4) =
=30.2 m/s
The time for this motion is
t₂=V₀/g = 30.2/9.8=3.1 s
Therefore, the apple has to start after Δt=t₁-t₂=5.2-3.1 =2.1 s.
(a) Well, well, looks like we have an apple that wants to join the archery party! To find the initial speed of the apple, we need to consider the vertical motion of the arrow and the horizontal motion of the apple.
The vertical component of the arrow's velocity is given by Vay = V * sin(theta), where V is the arrow's velocity (45 m/s) and theta is the launch angle (50 degrees). Since we want the apple to meet the path of the arrow, the vertical component of the apple's initial speed must be the same as Vay.
So, Vay_apple = V * sin(theta).
Now, let's figure out the time it takes for the arrow to reach the assistant's location, which is 150m away from the launch point. We can use the horizontal component of the arrow's velocity to find the time of flight.
The horizontal component of the arrow's velocity is Vax = V * cos(theta), and the time of flight is given by t = d / Vax, where d is the horizontal distance traveled (150m).
So, t = 150m / (V * cos(theta)).
Now that we know the time it takes for the arrow to reach the assistant's location, we can find the apple's initial speed by multiplying it by the time it takes for the arrow to reach the assistant's location:
V_apple = Vay_apple * t.
And voila, we have the initial speed of the apple!
(b) To find the time at which the apple should be thrown so that the arrow hits it, we know that the apple and the arrow must meet at the same horizontal position at the same time.
Since we already know the time it takes for the arrow to reach the assistant's location (t), we can simply add this time to the total time of flight of the apple.
The total time of flight of the apple is given by T_apple = 2 * V_apple / g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, the time after the arrow launch at which the apple should be thrown is t + T_apple.
Now, grab your juggling pins and have some fun making that apple meet the arrow!
To find the initial speed of the apple, we need to first calculate the horizontal and vertical components of the arrow's velocity.
The horizontal component of the arrow's velocity is given by:
Vx = V * cos(theta)
where Vx is the horizontal component of the velocity, V is the velocity of the arrow, and theta is the angle with the horizontal.
Vx = 45 m/s * cos(50 degrees)
Vx = 45 m/s * 0.6428
Vx = 28.92 m/s
Now, let's calculate the time it takes for the arrow to reach the assistant, who is standing 150 m downrange.
We can calculate the time using the horizontal distance and the horizontal component of velocity.
Time = distance / velocity
Time = 150 m / 28.92 m/s
Time = 5.19 seconds
Therefore, it takes 5.19 seconds for the arrow to reach the assistant.
To determine the minimum initial speed of the apple, we need to consider that the apple and arrow should meet at the same height and time.
The vertical component of the arrow's velocity is given by:
Vy = V * sin(theta)
where Vy is the vertical component of the velocity, V is the velocity of the arrow, and theta is the angle with the horizontal.
Vy = 45 m/s * sin(50 degrees)
Vy = 45 m/s * 0.7660
Vy = 34.47 m/s
Since the apple is thrown straight up, the initial vertical velocity of the apple is equal to the final vertical velocity of the arrow.
To find the minimum initial speed of the apple, we need to consider that the arrow and apple both reach their maximum height at the same time. The time it takes both the arrow and apple to reach their maximum height is half of the total time it takes the arrow to reach the assistant.
Therefore, the initial vertical velocity of the apple can be calculated as follows:
Vy_apple = Vy_arrow / 2
Vy_apple = 34.47 m/s / 2
Vy_apple = 17.24 m/s
Now, we can find the initial speed of the apple using the horizontal and vertical components of velocity:
Initial speed of the apple = sqrt((Vx)^2 + (Vy_apple)^2)
Initial speed of the apple = sqrt((28.92 m/s)^2 + (17.24 m/s)^2)
Initial speed of the apple = sqrt(837.5568 + 297.6576)
Initial speed of the apple = sqrt(1135.2144)
Initial speed of the apple = 33.69 m/s
(a) Therefore, the minimum initial speed of the apple is approximately 33.69 m/s.
(b) To find the time at which the apple should be thrown so that the arrow hits the apple, we need to consider that the apple should be thrown at the same time the arrow reaches the assistant, which is 5.19 seconds after launch.
Therefore, the apple should be thrown 5.19 seconds after the arrow is launched.
To solve this problem, we need to break it down into two parts: (a) determining the initial speed of the apple, and (b) finding the time when the apple should be thrown. Let's solve each part step by step.
(a) To find the initial speed of the apple, we can use the concept of projectile motion. Both the arrow and the apple are projectiles, and they follow a parabolic path. The vertical motion of the apple must be the same as the vertical motion of the arrow when they intersect.
First, let's find the time it takes for the arrow to reach the horizontal distance of 150m. We can use the horizontal component of the arrow's velocity:
Horizontal velocity (Vx) = Velocity × cos(angle)
Vx = 45 m/s × cos(50°)
Vx = 45 m/s × 0.64279
Vx = 28.92455 m/s
To find the time (t) it takes for the arrow to reach the target,
Horizontal distance (dx) = Horizontal Velocity × Time
150m = 28.92455 m/s × t
Solving for t:
t = 150m / 28.92455 m/s
t ≈ 5.18 seconds
Now, let's consider the vertical motion of the apple. When the apple and the arrow intersect, they will have the same vertical displacement (let's call it dy). The vertical displacement of the apple can be determined using the equation:
dy = Vya × t + 0.5 × g × t²
Where:
Vya is the initial vertical velocity of the apple
t is the time taken for the arrow to reach the target (5.18 seconds)
g is the acceleration due to gravity (9.8 m/s²)
Since the apple is thrown straight up, the vertical component of its initial velocity Vya will be negative (opposite to the gravitational acceleration):
dy = Vya × t - 0.5 × g × t²
Since the arrow is launched at an angle of 50 degrees, the vertical displacement dy can be determined using the equation:
dy = 45 m/s × sin(50°) × t - 0.5 × 9.8 m/s² × t²
Now, we can set this equation equal to zero, as the apple has to meet the path of the arrow when they intersect:
0 = 45 m/s × sin(50°) × t - 0.5 × 9.8 m/s² × t²
Simplifying the equation:
0 = 22.5 m/s × t - 4.9 m/s² × t²
This is a quadratic equation in terms of t. To solve for t, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Where:
a = -4.9 m/s²
b = 22.5 m/s
c = 0
Solving for t:
t = (-22.5 ± √(22.5² - 4(-4.9)(0))) / (2(-4.9))
t = (-22.5 ± √(506.25)) / (-9.8)
Taking the positive root:
t = (-22.5 + √(506.25)) / (-9.8)
t ≈ 2.03 seconds
Therefore, the initial speed of the apple is approximately 2.03 seconds.
(b) Since we have already determined that the time taken for the arrow to reach the target is approximately 5.18 seconds, and the apple must be thrown at the same time, the assistant should throw the apple 5.18 seconds after the arrow is launched.
So, to summarize:
(a) The initial speed of the apple is approximately 2.03 m/s.
(b) The apple should be thrown 5.18 seconds after the arrow is launched.