A shell leaves a mortar with a muzzle velocity of 500 ft/s directed upward at 60 degrees with the horizontal. Determine the position of the shell and its resultant velocity 20 s after firing. how high will it rise?
Correction: V = Xo + Yi = 250 - 207i = 325Ft/s[-39.6o].
Vo = 500Ft/s[60o].
Xo = 500*Cos60 = 250 Ft/s.
Yo = 500*sin60 = 433 Ft/s.
Y = Yo + g*Tr = 0, 433 - 32Tr = 0, Tr = 13.53 s. Rise time.
h = Yo*t + 0.5g*t^2.
h = 433*20 - 16*20^2 = 2260 Ft.
and falling.
Y = Yo + g*t = 433 - 32*20 = -207 Ft/s = 207 Ft/s, Downward.
V = Xo + Yi = 250 - .53= 325 Ft/s[-39.6o].
h max = Yo*Tr + 0.5g*Tr^2.
h max = 433*13.53 - 16*13.53^2 = 2930 Ft.
To determine the position and resultant velocity of the shell 20 seconds after firing, we first need to break down the initial velocity into its horizontal and vertical components.
Given:
Muzzle velocity (initial velocity), v₀ = 500 ft/s
Launch angle, θ = 60 degrees
Step 1: Determine the initial horizontal and vertical velocity components.
The horizontal component of velocity (v₀x) is given by:
v₀x = v₀ * cos(θ)
The vertical component of velocity (v₀y) is given by:
v₀y = v₀ * sin(θ)
Substituting the values:
v₀x = 500 ft/s * cos(60°)
v₀x ≈ 250 ft/s
v₀y = 500 ft/s * sin(60°)
v₀y ≈ 433 ft/s
Step 2: Determine the position after 20 seconds.
The horizontal distance traveled (x) is given by:
x = v₀x * t
Substituting the values:
x = 250 ft/s * 20 s
x = 5000 ft
The vertical distance traveled (y) is determined by considering the free-fall motion:
y = v₀yt - (1/2) * g * t²
where g is the acceleration due to gravity, which is approximately 32.2 ft/s² (on Earth) in this case.
Substituting the values:
y = (433 ft/s) * 20 s - (1/2) * 32.2 ft/s² * (20 s)²
y = 8660 ft - 3220 ft
y ≈ 5440 ft
Step 3: Determine the maximum height reached.
The maximum height (H) reached by the shell can be calculated as:
H = (v₀y)² / (2 * g)
Substituting the values:
H = (433 ft/s)² / (2 * 32.2 ft/s²)
H ≈ 297.2 ft
Therefore, the shell will reach a height of approximately 297.2 ft.
Step 4: Determine the resultant velocity.
The resultant velocity (v) at any given time (20 s in this case) can be determined using the horizontal and vertical components.
The horizontal component (vx) remains constant throughout. Therefore:
vx = v₀x = 250 ft/s
The vertical component (vy) changes due to the influence of gravity. At any given time (t), it is given by:
vy = v₀y - g * t
Substituting the values:
vy = 433 ft/s - 32.2 ft/s² * 20 s
vy = 433 ft/s - 644 ft/s
vy ≈ -211 ft/s (negative sign indicates downward)
The resultant velocity (v) is given by:
v = √(vx² + vy²)
Substituting the values:
v = √(250 ft/s)² + (-211 ft/s)²
v ≈ √(62500 ft²/s² + 44521 ft²/s²)
v ≈ √107021 ft²/s²
v ≈ 327 ft/s (rounded to the nearest whole number)
Therefore, 20 seconds after firing:
- The position of the shell is approximately 5000 ft horizontally and 5440 ft vertically above the point of firing.
- The resultant velocity of the shell is approximately 327 ft/s, directed downward.
To determine the position of the shell and its resultant velocity, we can break down the problem into two components: horizontal and vertical.
1. Horizontal Component:
The horizontal component of the velocity remains constant throughout the flight because there is no acceleration acting in the horizontal direction. So, the horizontal component of the velocity remains the same as the muzzle velocity of 500 ft/s.
2. Vertical Component:
The vertical component of the velocity changes due to the acceleration due to gravity acting downward. We can analyze the vertical motion by using the equations of motion.
a) First, we need to determine the initial vertical velocity (Vy0). We can use trigonometry to calculate this:
Vy0 = V * sin(θ)
where V is the muzzle velocity (500 ft/s) and θ is the angle with the horizontal (60 degrees).
Vy0 = 500 ft/s * sin(60 degrees)
Vy0 = 500 ft/s * 0.866
Vy0 ≈ 433 ft/s
b) Now, let's determine the time it takes for the shell to reach its highest point (t_max) using the equation:
t_max = Vy0 / g
where g is the acceleration due to gravity (32.2 ft/s^2).
t_max = 433 ft/s / 32.2 ft/s^2
t_max ≈ 13.45 s
c) Next, we can find the maximum height (H) reached by the shell using the equation of motion:
H = Vy0 * t_max - (1/2) * g * t_max^2
H = 433 ft/s * 13.45 s - (1/2) * 32.2 ft/s^2 * (13.45 s)^2
H ≈ 2914.5 ft
So, the shell will rise to a height of approximately 2914.5 ft.
d) Finally, we can determine the position of the shell and its resultant velocity 20 seconds after firing.
Position: The horizontal position will be given by the formula:
X = Vx * t
where Vx is the horizontal component of the velocity and t is the time.
X = 500 ft/s * 20 s
X = 10000 ft
So, the shell will be 10000 ft away from the launching point in the horizontal direction.
Resultant Velocity: The resultant velocity will be the vector sum of the horizontal and vertical velocities at any given time. Since there is no horizontal acceleration, the horizontal component remains constant, and the vertical component decreases due to the acceleration of gravity.
After 20 seconds, the vertical component of the velocity will be:
Vy = Vy0 - g * t
Vy = 433 ft/s - 32.2 ft/s^2 * 20 s
Vy ≈ 433 ft/s - 644 ft/s
Vy ≈ -211 ft/s
The negative sign indicates that the velocity is directed downward.
The resultant velocity (Vr) can be calculated using Pythagoras' theorem:
Vr = sqrt(Vx^2 + Vy^2)
Vr = sqrt((500 ft/s)^2 + (-211 ft/s)^2)
Vr ≈ sqrt(250000 ft^2/s^2 + 44521 ft^2/s^2)
Vr ≈ sqrt(294521 ft^2/s^2)
Vr ≈ 542 ft/s
So, the resultant velocity of the shell 20 seconds after firing is approximately 542 ft/s downward.
In summary:
- The vertical position of the shell 20 seconds after firing will be approximately 2914.5 ft above the ground.
- The horizontal position of the shell will be 10000 ft away from the launching point.
- The resultant velocity of the shell 20 seconds after firing will be approximately 542 ft/s downward.