3) A shell is fired from ground level with a muzzle speed of 350 ft / s and elevation and angle
of 60 . Find
(a) parametric equation for the shell’s trajectory
(b) the maximum height reached by the shell
(c) the horizontal distance traveled by the shell
(d) the speed of the shell at impact.
Vo = 350Ft/s[60o].
Xo = 350*cos60 = 175 Ft/s
Yo = 350*sin60 = 303 Ft/s.
b. Y^2 = Yo^2 + 2g*h
h = (Y^2-Yo^2)/2g
h = (0-(303)^2)/-64 = 1435 Ft.
c. Dx = Vo^2*sin(2A)/g
Dx = (350)^2*sin(120)/32 = 3315 Ft.
d. 350 Ft/s[60o]. Same as initial velocity.
(a) The parametric equation for the shell's trajectory can be expressed as follows:
x(t) = v0 * cos(θ) * t
y(t) = v0 * sin(θ) * t - (1/2) * g * t^2
Where:
- x(t) is the horizontal position of the shell at time t
- y(t) is the vertical position of the shell at time t
- v0 is the muzzle speed of the shell (350 ft/s)
- θ is the launch angle (60 degrees)
- g is the acceleration due to gravity (32.2 ft/s^2)
(b) To find the maximum height reached by the shell, we need to find the time at which it reaches its peak. At the highest point, the vertical velocity component is zero.
v_y(t) = v0 * sin(θ) - g * t
Setting v_y(t) = 0 and solving for t, we get:
t = v0 * sin(θ) / g
Substituting the given values:
t = (350 ft/s) * sin(60 degrees) / 32.2 ft/s^2
Now we can find the maximum height by substituting t into the y(t) equation:
max height = y(t) = (350 ft/s) * sin(60 degrees) * [(350 ft/s) * sin(60 degrees) / 32.2 ft/s^2] - (1/2) * 32.2 ft/s^2 * [(350 ft/s) * sin(60 degrees) / 32.2 ft/s^2]^2
(c) The horizontal distance traveled by the shell can be determined by finding the time it takes for the shell to reach the ground. This occurs when y(t) equals zero:
0 = v0 * sin(θ) * t - (1/2) * g * t^2
Solving for t, we get:
t = [v0 * sin(θ)] / g
Substituting the given values:
t = (350 ft/s) * sin(60 degrees) / 32.2 ft/s^2
Now we can find the horizontal distance by substituting t into the x(t) equation:
horizontal distance = x(t) = (350 ft/s) * cos(60 degrees) * [(350 ft/s) * sin(60 degrees) / 32.2 ft/s^2]
(d) To find the speed of the shell at impact, we can use the x(t) equation to determine the time it takes for the shell to reach the ground and then substitute this value into the equation for the horizontal velocity.
t_impact = [v0 * sin(θ)] / g
Substituting the given values:
t_impact = (350 ft/s) * sin(60 degrees) / 32.2 ft/s^2
Now we can find the speed at impact:
speed at impact = x'(t_impact) = v0 * cos(θ)
Substituting the given values:
speed at impact = (350 ft/s) * cos(60 degrees)
Please note that the numerical values given in the explanation may not be accurate, as they are used for illustration purposes only.
To find the parametric equation for the shell's trajectory, we can use the equations of motion for projectile motion. The equations are:
Horizontal motion:
x(t) = v0 * cos(θ) * t
Vertical motion:
y(t) = v0 * sin(θ) * t - (1/2) * g * t^2
where:
v0 = initial velocity = 350 ft/s
θ = angle = 60 degrees
g = acceleration due to gravity = 32 ft/s^2
(a) Parametric equation for the shell's trajectory:
Taking time t as the parameter, the parametric equations become:
x(t) = v0 * cos(θ) * t
y(t) = v0 * sin(θ) * t - (1/2) * g * t^2
(b) To find the maximum height reached by the shell, we need to determine the time at which the vertical component of velocity becomes zero. This happens when:
v_y(t) = v0 * sin(θ) - g * t = 0
t = v0 * sin(θ) / g
Substituting this value of t into the equation for height, we get:
y_max = v0 * sin(θ) * t - (1/2) * g * t^2
(c) The horizontal distance traveled by the shell can be found by determining the time at which the shell hits the ground. This occurs when y(t) = 0. Solving the equation for t, we get:
v0 * sin(θ) * t - (1/2) * g * t^2 = 0
t = 2 * v0 * sin(θ) / g
Substituting this value of t into the equation for horizontal distance, we have:
x_distance = v0 * cos(θ) * t
(d) To find the speed of the shell at impact, we can use the horizontal velocity, given by:
v_x = v0 * cos(θ)
Since the horizontal velocity remains constant throughout the motion, the speed of the shell at impact is 350 ft/s.
To solve this problem, we can break it down into several steps. Let's start with part (a) and find the parametric equation for the shell's trajectory.
(a) Parametric Equation for the Shell's Trajectory:
The parametric equations for a projectile in 2D motion can be given as:
x(t) = v₀ * cos(θ) * t
y(t) = v₀ * sin(θ) * t - (1/2) * g * t²
Where:
x(t) represents the horizontal distance traveled by the shell at time t
y(t) represents the vertical distance (height) of the shell at time t
v₀ represents the initial velocity (muzzle speed) of the shell
θ represents the launch angle
g represents the acceleration due to gravity (approximately 9.8 m/s²)
In this case, we are given the muzzle speed (v₀ = 350 ft/s) and the launch angle (θ = 60°). However, it's important to note that the equations above assume the use of SI units, so we need to convert the given muzzle speed from feet per second (ft/s) to meters per second (m/s).
Using the conversion factor, 1 ft = 0.3048 m, the muzzle speed in meters per second is:
v₀ = 350 ft/s * 0.3048 m/ft ≈ 106.68 m/s
Now we can plug in the values into the parametric equations:
x(t) = 106.68 m/s * cos(60°) * t
y(t) = 106.68 m/s * sin(60°) * t - (1/2) * 9.8 m/s² * t²
Simplifying the equations:
x(t) = 53.34t
y(t) = 58.52t - 4.9t²
Therefore, the parametric equations for the shell's trajectory are x(t) = 53.34t and y(t) = 58.52t - 4.9t².
Now let's move on to part (b) and find the maximum height reached by the shell.
(b) Maximum Height Reached by the Shell:
The maximum height can be determined by finding the vertex of the parabolic trajectory. In this case, since the equation for the vertical distance of the shell is y(t) = 58.52t - 4.9t², we can set the derivative of y(t) with respect to t equal to zero to find the time (t) when the height is maximum.
dy/dt = 58.52 - 9.8t = 0
Solving for t:
58.52 - 9.8t = 0
9.8t = 58.52
t = 58.52 / 9.8 ≈ 5.97 seconds
Now we can plug this value of t back into the equation for y(t) to find the maximum height:
y(t) = 58.52 * 5.97 - 4.9 * (5.97)²
Calculating the height:
y(t) ≈ 352.77 meters
Therefore, the maximum height reached by the shell is approximately 352.77 meters.
Next, we'll move on to part (c) and find the horizontal distance traveled by the shell.
(c) Horizontal Distance Traveled by the Shell:
The horizontal distance traveled by the shell can be determined by finding the time it takes for the shell to reach the ground. In this case, we can set y(t) = 0 (since the shell will be on the ground) and solve for t.
0 = 58.52t - 4.9t²
Simplifying the equation:
4.9t² - 58.52t = 0
t(4.9t - 58.52) = 0
Now we have two possible solutions for t:
t₁ = 0 (initial time when the shell is launched)
t₂ = 58.52 / 4.9 = 11.95 seconds
We can ignore t = 0 since it represents the time when the shell is launched. Therefore, the time it takes for the shell to reach the ground is approximately 11.95 seconds.
Now we can plug this value of t back into the equation for x(t) to find the horizontal distance:
x(t) = 53.34 * 11.95
Calculating the distance:
x(t) ≈ 637.01 meters
Therefore, the horizontal distance traveled by the shell is approximately 637.01 meters.
Finally, let's move on to part (d) and find the speed of the shell at impact.
(d) Speed of the Shell at Impact:
The speed of the shell at impact can be determined by calculating the magnitude of its velocity vector at the time it hits the ground.
The magnitude of the velocity vector can be found by taking the square root of the sum of the squares of its horizontal and vertical components:
vx = dx/dt = 53.34 m/s (constant throughout the trajectory)
vy = dy/dt = 58.52 - 9.8t
Using the time t = 11.95 seconds (time at impact), we can calculate the vertical component of the velocity at impact:
vy = 58.52 - 9.8 * 11.95
Calculating the vertical component:
vy ≈ -103.953 m/s
Now we can calculate the magnitude of the velocity vector:
v_impact = √(vx² + vy²)
Substituting the values:
v_impact = √((53.34)² + (-103.953)²)
Calculating the magnitude:
v_impact ≈ 118.05 m/s
Therefore, the speed of the shell at impact is approximately 118.05 m/s.
To summarize:
(a) The parametric equations for the shell's trajectory are x(t) = 53.34t and y(t) = 58.52t - 4.9t².
(b) The maximum height reached by the shell is approximately 352.77 meters.
(c) The horizontal distance traveled by the shell is approximately 637.01 meters.
(d) The speed of the shell at impact is approximately 118.05 m/s.