A cannon fires a 0.639 kg shell with initial
velocity vi = 10 m/s in the direction θ = 55
◦
above the horizontal.
The shell’s trajectory curves downward because of gravity, so at the time t = 0.217 s
the shell is below the straight line by some
vertical distance ∆h.
Find this distance ∆h in the absence of
air resistance. The acceleration of gravity is
9.8 m/s
2
.
Answer in units of m
To find the vertical distance ∆h in the absence of air resistance, we can use the equations of motion.
First, we need to find the vertical displacement (Δy) at time t = 0.217s when the shell is below the straight line.
Using the equation:
Δy = viy * t + (1/2) * a * t^2
where:
- viy is the vertical component of the initial velocity
- a is the acceleration due to gravity
We can break down the initial velocity (vi) into its horizontal and vertical components.
The vertical component can be found by:
viy = vi * sin(θ)
Given:
vi = 10 m/s
θ = 55 degrees
viy = 10 * sin(55)
Next, we need to substitute the known values into the equation to find Δy.
Using:
t = 0.217s
a = 9.8 m/s^2
Δy = (viy * t) + (1/2) * a * t^2
Now we can solve for Δy.
Δy = (viy * t) + (1/2) * a * t^2
Δy = (10 * sin(55) * 0.217) + (1/2) * 9.8 * (0.217)^2
Calculate the value of sin(55) and substitute it into the equation above:
sin(55) = 0.819
Δy = (10 * 0.819 * 0.217) + (1/2) * 9.8 * (0.217)^2
Δy = 1.780 + (0.5) * 9.8 * 0.047089
Δy = 1.780 + 0.230439
Δy = 2.010439
Therefore, the vertical distance Δh in the absence of air resistance is approximately 2.010 m.
To find the vertical distance ∆h traveled by the shell in the absence of air resistance, we can use the equations of projectile motion.
First, we need to break down the initial velocity into its horizontal and vertical components. The horizontal component (vx) remains constant as there is no acceleration in that direction. The vertical component (vy) changes due to the acceleration of gravity.
Given:
Initial velocity (vi) = 10 m/s
Launch angle (θ) = 55°
Time (t) = 0.217 s
Acceleration due to gravity (g) = 9.8 m/s^2
1. Calculate the horizontal component of velocity:
vx = vi * cos(θ)
vx = 10 * cos(55°)
vx ≈ 5.108 m/s
2. Calculate the vertical component of velocity:
vy = vi * sin(θ)
vy = 10 * sin(55°)
vy ≈ 8.66 m/s
3. Calculate the displacement in the horizontal direction:
The displacement in the horizontal direction can be found using the formula:
Δx = vx * t
Δx = 5.108 m/s * 0.217 s
Δx ≈ 1.108 m
4. Calculate the displacement in the vertical direction:
The displacement in the vertical direction can be found using the formula:
Δy = vy * t + (1/2) * g * t^2
Δy = 8.66 m/s * 0.217 s + (1/2) * 9.8 m/s^2 * (0.217 s)^2
Δy ≈ 1.856 m
5. Calculate the vertical distance ∆h:
Since the trajectory curves downward, the vertical distance ∆h is equal to -Δy. Hence,
∆h = -1.856 m (approximately)
Therefore, the vertical distance traveled by the shell, ∆h, in the absence of air resistance is approximately -1.856 m (with a negative sign indicating it is below the straight line).
Llkkk
Mistake: it has to be “above the straight line”
Δh=vₒ(y) •t-gt²/2= vₒ•sinα•t- gt²/2=
=10•sin55•0.217 – 9.8•0.217²/2 =
= 1.78 – 0.23 =1.55 m