A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wall to break apart the wall. The machine was not placed near the wall because then arrows could reach it from the castle wall. Instead, it was positioned so that the stone hit the wall during the second half of its flight. Suppose a stone is launched with a speed of v0 = 27.0 m/s and at an angle of θ0 = 42.0°. What is the speed of the stone if it hits the wall (a) just as it reaches the top of its parabolic path and (b) when it has descended to half that height? (c) As a percentage, how much faster is it moving in part (b) than in part (a)?
I don't understand why we made vy^2 to be zero first to find the height.can u please explain
To solve this problem, we need to break it down into several steps.
Step 1: Calculate the initial horizontal and vertical components of the stone's velocity.
The initial horizontal component of velocity (v0x) can be calculated using the formula:
v0x = v0 * cos(θ0)
where v0 is the initial speed and θ0 is the launch angle.
v0x = 27.0 m/s * cos(42.0°)
v0x ≈ 20.07 m/s
The initial vertical component of velocity (v0y) can be calculated using the formula:
v0y = v0 * sin(θ0)
v0y = 27.0 m/s * sin(42.0°)
v0y ≈ 18.14 m/s
Step 2: Calculate the time it takes for the stone to reach the top of its parabolic path.
The time it takes for an object to reach its maximum height can be calculated using the formula:
t = v0y / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
t = 18.14 m/s / 9.8 m/s²
t ≈ 1.85 s
Step 3: Calculate the maximum height reached by the stone.
The maximum height (H) reached by the stone can be calculated using the formula:
H = (v0y^2) / (2 * g)
H = (18.14 m/s)^2 / (2 * 9.8 m/s²)
H ≈ 16.51 m
Step 4: Calculate the horizontal component of the stone's velocity when it reaches the top of its path (part a).
The horizontal component of velocity remains constant throughout the flight, so the horizontal component of velocity at any point in the trajectory is equal to the initial horizontal component of velocity (v0x = 20.07 m/s).
Step 5: Calculate the vertical component of velocity when the stone has descended to half the maximum height (part b).
The vertical component of velocity (v1y) at any point in the trajectory can be calculated using the formula:
v1y = v0y - g * t
where t is the time taken to reach that point.
t1 = t / 2
t1 ≈ 0.93 s
v1y = 18.14 m/s - 9.8 m/s² * 0.93 s
v1y ≈ 9.33 m/s
Step 6: Calculate the speed of the stone when it reaches the top of its path (part a).
The speed of the stone at any point in the trajectory can be calculated using the formula:
v = √(v_x^2 + v_y^2)
where v_x is the horizontal component of velocity and v_y is the vertical component of velocity.
v = √((v0x)^2 + (v0y)^2)
v ≈ √((20.07 m/s)^2 + (18.14 m/s)^2)
v ≈ 26.48 m/s
Step 7: Calculate the speed of the stone when it has descended to half the maximum height (part b).
v1 = √((v0x)^2 + (v1y)^2)
v1 ≈ √((20.07 m/s)^2 + (9.33 m/s)^2)
v1 ≈ 22.39 m/s
Step 8: Calculate the percentage difference in speed between part a and part b.
The percentage difference can be calculated using the formula:
percentage difference = ((v1 - v) / v) * 100
percentage difference = ((22.39 m/s - 26.48 m/s) / 26.48 m/s) * 100
percentage difference ≈ -15.49%
Therefore, the stone is moving approximately 15.49% slower in part (b) than in part (a).
To determine the speed of the stone in different scenarios, we can use the principles of projectile motion. Let's break down the problem into three parts: (a) just as it reaches the top of its parabolic path, (b) when it has descended to half that height, and (c) the percentage difference between (a) and (b).
(a) Just as it reaches the top of its parabolic path:
At the highest point of a projectile's trajectory, the vertical component of its velocity is zero. Therefore, we only need to consider the horizontal component of velocity and calculate its magnitude.
To find the horizontal velocity (v_x) at any point, we can use the formula:
v_x = v₀ * cos θ₀
where:
v₀ = 27.0 m/s (initial velocity)
θ₀ = 42.0° (launch angle)
Substituting the known values into the equation:
v_x = 27.0 m/s * cos(42.0°)
Calculating v_x:
v_x ≈ 20.474 m/s
Since the stone will have the same horizontal velocity throughout its flight, the speed of the stone when it hits the wall during the second half of its flight (a) will be equal to v_x:
Speed of the stone when it hits the wall at the top of its path (a) ≈ 20.474 m/s
(b) When it has descended to half that height:
To determine the speed of the stone when it has descended to half of its maximum height, we need to calculate the new height (h).
Since the stone follows a symmetric parabolic trajectory, the height at half the maximum height is half the maximum height.
Let's assume the maximum height is represented by H, then:
h = H/2
Next, we can use the following equation for the height of the projectile at any given time:
h = v₀^2 * (sin θ₀)^2 / (2 * g)
where:
v₀ = 27.0 m/s (initial velocity)
θ₀ = 42.0° (launch angle)
g = 9.8 m/s² (acceleration due to gravity)
Substituting the known values into the equation:
h = (27.0 m/s)^2 * (sin 42.0°)^2 / (2 * 9.8 m/s²)
Calculating h:
h ≈ 17.711 m
Now that we have the new height, we need to determine the speed of the stone when it has descended to this height. To calculate the new velocity, we need to find the vertical component of velocity (v_y) when h = 17.711 m.
We can use the following equation for the vertical component of velocity at any given time:
v_y = v₀ * sin θ₀ - g * t
where:
v₀ = 27.0 m/s (initial velocity)
θ₀ = 42.0° (launch angle)
g = 9.8 m/s² (acceleration due to gravity)
t = time taken to reach the desired height
At the top of the trajectory, the vertical velocity is zero, and at the desired height, half the maximum height, the time taken can be obtained using the following equation:
h = v₀ * sin θ₀ * t - (1/2) * g * t^2
Rearranging the equation and substituting the known values:
(1/2) * 17.711 m = (27.0 m/s * sin 42.0° * t) - (1/2) * 9.8 m/s² * t^2
Simplifying and solving the equation for t:
4.855 m = (22.314 m/s * t) - 4.9 m/s² * t^2
4.9 m/s² * t^2 - 22.314 m/s * t + 4.855 m = 0
Solving the quadratic equation for t, we find:
t ≈ 0.890 s
Now that we know t, let's calculate the vertical component of velocity (v_y):
v_y = 27.0 m/s * sin 42.0° - 9.8 m/s² * 0.890 s
Calculating v_y:
v_y ≈ 16.201 m/s
Finally, we can find the speed of the stone when it hits the wall at half of its maximum height (b) by calculating the resultant velocity (v):
v = √(v_x^2 + v_y^2)
Substituting the known values into the equation:
v = √((20.474 m/s)^2 + (16.201 m/s)^2)
Calculating v:
v ≈ 26.046 m/s
Speed of the stone when it hits the wall at half of the maximum height (b) ≈ 26.046 m/s
(c) Percentage difference in speed between (a) and (b):
To find the percentage difference between the speeds in parts (a) and (b), we can use the following formula:
Percentage difference = (|v_b - v_a| / v_a) * 100%
Substituting the known values into the equation:
Percentage difference = (|26.046 m/s - 20.474 m/s| / 20.474 m/s) * 100%
Calculating the percentage difference:
Percentage difference ≈ 26.97%
Therefore, the stone is moving approximately 26.97% faster in part (b) than in part (a).
Vo = 27m/s[42o].
Xo = 27*Cos42 = 20.1 m/s.
Yo = 27*sin42 = 18.1 m/s.
a. Y = 0, V = Xo = 20.1 m/s.
b. Y^2 = Yo^2 + 2g*h = 0.
h = -Yo^2/2g = -(18.1^2)/-19.6 = 16.7 m, max.
Y^2 = Yo^2 + 2g*h/2. = 0 + 19.6*8.35 =
163.66.
Y = 12.8 m/s.
V = Sqrt(Xo^2 + Y^2).
c. (Xo+Y) - Xo = (20.1+12.8i) - 20.1 =
12.8i = 12.8m/s[90o].