An archer shoots an arrow with a velocity of 49.5 m/s at an angle of 52.0¡Æ with the horizontal. An assistant standing on the level ground 150 m downrange from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow.
(a) What is the initial speed of the apple?
m/s
(b) At what time after the arrow launch should the apple be thrown so that the arrow hits the apple?
a. Vo = 49.5 m/s @ 52o.
Xo = 49.5*cos52 = 30.5 m/s.
Yo = 49.5*sin52 = 39.0 m/s.
hmax = (Y^2-Yo^2)/2g.
hmax = (0-1521) / -19.6 = 77.6 m.
Tr = (Y-Yo)/g.
Tr = (0-39 / -9.8 = 3.98 s. = Rise time.
Xo*t = Dx.
t = Dx / Xo = 150 / 30.5 = 4.92 s.=Time
to reach 150 m range.
h = ho - 0.5g*t^2
h = 77.6 - 4.9*(4.92-3.98)^2 = 73.3 m.=
Dist. from arrow to gnd @ 150 m range.
Vo^2 + 2g*h = V^2.
Vo^2 - 19.6*73.3 = 0
Vo^2 - 1436.68
Vo=37.9 m/s.=Initial velocity of apple.
b. Tr = (V-Vo)/g.
Tr = (0-37.9) / -9.8 = 3.87 s. = Rise
time of apple.
T = 4.92 - 3.87 = 1.05 s after arrow is launched
To solve this problem, we can break it down into two parts:
Part 1: Finding the initial speed of the apple (Vertical Motion)
In this part, we need to consider the vertical motion of the apple. The arrow is launched at an angle of 52.0 degrees, so we can find the vertical component of its velocity.
1. Resolve the arrow's initial velocity into its vertical and horizontal components:
Vertical component (Vay) = Arrow's initial velocity (49.5 m/s) * sin(angle)
= 49.5 * sin(52.0 degrees)
2. The apple needs to reach the same height as the arrow at the same time, so the time taken by the apple to reach its maximum height should be half of the total time it takes for the arrow to travel downrange (since we ignore air resistance).
3. Use the kinematic equation to find the time taken by the apple to reach its maximum height:
Vertical displacement = (Vay)^2 / (2 * acceleration)
0 = (Vay)^2 / (2 * acceleration)
acceleration = 9.8 m/s^2 (acceleration due to gravity)
Substitute the known values and solve for time:
0 = (Vay)^2 / (2 * 9.8)
(Vay)^2 = 19.6
Vay = sqrt(19.6)
4. The initial velocity of the apple is equal to the vertical component of velocity, so the answer to part (a) is:
Initial speed of the apple = Vay
= sqrt(19.6)
Part 2: Finding the time for the arrow to reach the assistant (Horizontal Motion)
In this part, we need to consider the horizontal motion of the arrow and find the time it takes to reach the assistant.
1. Resolve the arrow's initial velocity into its vertical and horizontal components:
Horizontal component (Vax) = Arrow's initial velocity (49.5 m/s) * cos(angle)
= 49.5 * cos(52.0 degrees)
2. Use the formula for horizontal motion to find the time taken by the arrow to reach the assistant:
Distance = Vax * time
150 m = (49.5 * cos(52.0 degrees)) * time
Solve for time:
time = 150 / (49.5 * cos(52.0 degrees))
3. The apple needs to be thrown at the same time as the arrow reaches the assistant. Therefore, the answer to part (b) is:
Time after the arrow launch = time
Now you can plug in the known values and calculate the answers.
To solve this problem, we can break it down into two parts:
Part 1: Find the initial speed of the apple
To find the minimum initial speed of the apple, we need to consider the horizontal and vertical components of motion separately.
Horizontal motion:
The horizontal component of the apple's initial velocity will be the same as the arrow's velocity, as it needs to meet the arrow's path.
Given:
Arrow's velocity = 49.5 m/s
Vertical motion:
To find the initial speed of the apple's vertical motion, we can use the equation for projectile motion:
y = voy * t + (1/2) * g * t^2
where:
y = 150 meters (distance downrange)
voy = initial vertical velocity of the apple in m/s
t = time taken for the apple to reach 150 meters
g = acceleration due to gravity (approximately 9.8 m/s^2)
Since the apple is thrown straight up, its final vertical velocity will be zero when it reaches its highest point. Therefore, the equation becomes:
0 = voy * t + (1/2) * g * t^2
Using this equation, we can solve for the initial vertical velocity, voy.
Part 2: Find the time at which the apple should be thrown
Now that we know the initial vertical velocity, voy, we can use it to find the time at which the apple should be thrown to meet the arrow's path.
We know that the horizontal component of the apple's velocity is the same as the arrow's velocity, which is 49.5 m/s.
Using the equation:
x = vox * t
where:
x = 150 meters (distance downrange)
vox = initial horizontal velocity of the apple in m/s
t = time taken for the apple to reach 150 meters
We can solve for t to find the time at which the apple should be thrown.
Let's solve for the answers step-by-step:
Part 1: Find the initial speed of the apple
Using the equation for vertical motion, we have:
0 = voy * t + (1/2) * g * t^2
Since the final vertical velocity is zero, the equation becomes:
(1/2) * g * t^2 = -voy * t
Rearranging the equation, we get:
voy = (-1/2) * g * t
Substituting the values:
g = 9.8 m/s^2
t = ? (unknown)
Part 2: Find the time at which the apple should be thrown
Using the equation for horizontal motion, we have:
x = vox * t
Since x = 150 meters and vox = 49.5 m/s, we can substitute these values into the equation:
150 = 49.5 * t
Now, let's solve each part step-by-step:
Part 1: Find the initial speed of the apple
Using the equation: voy = (-1/2) * g * t
Substitute the known values:
voy = (-1/2) * 9.8 * t
Part 2: Find the time at which the apple should be thrown
Using the equation: 150 = 49.5 * t
Now, let's solve each part step-by-step:
Part 1: Find the initial speed of the apple
Using the equation: voy = (-1/2) * g * t
Substitute the known values:
voy = (-1/2) * 9.8 * t
Part 2: Find the time at which the apple should be thrown
Using the equation: 150 = 49.5 * t
Solve for t:
t = 150 / 49.5
Simplifying, we get:
t ≈ 3.03 seconds
Now, let's substitute this value into the equation for voy:
voy = (-1/2) * 9.8 * t
voy = (-1/2) * 9.8 * 3.03
voy ≈ -14.9 m/s
We ignore the negative sign since it only indicates the direction of motion.
Therefore, the initial speed of the apple is approximately 14.9 m/s and should be thrown after approximately 3.03 seconds after the arrow is launched.