Assignment 4 Projectile Motions
A shell is fired from a cliff that is 30 m above a horizontal plane. The muzzle speed of the shell is 75.0 m/s and it is fired at an elevation of 35° above the horizontal.
(a) Determine the horizontal range of the shell.
(b) Determine the velocity of the shell as it strikes the ground.
To solve this problem, we can break down the motion of the shell into its horizontal and vertical components.
(a) To determine the horizontal range, we need to find the distance traveled by the shell horizontally. The horizontal component of the muzzle velocity can be found by multiplying the muzzle speed (75.0 m/s) by the cosine of the elevation angle (35°).
Horizontal component of velocity, Vx = 75.0 m/s * cos(35°)
Now, we need to find the time it takes for the shell to reach the ground. We can use the vertical component of the velocity to find the time of flight. The vertical component of the muzzle velocity can be found by multiplying the muzzle speed (75.0 m/s) by the sine of the elevation angle (35°).
Vertical component of velocity, Vy = 75.0 m/s * sin(35°)
The time of flight can be found using the equation: time = (2 * Vy) / g, where g is the acceleration due to gravity (9.8 m/s^2).
Time of flight, t = (2 * Vy) / g
Now, we can find the horizontal range using the equation: range = Vx * t
Substituting the values, we have:
Horizontal range = (75.0 m/s * cos(35°)) * [(2 * (75.0 m/s * sin(35°))) / 9.8 m/s^2]
(b) To determine the velocity of the shell as it strikes the ground, we can use the vertical component of velocity and the time of flight. The vertical component of the velocity at the time of impact can be found using the equation: Vf = Vy - g * t
Substituting the values, we have:
Vertical component of velocity at impact = 75.0 m/s * sin(35°) - (9.8 m/s^2) * [(2 * (75.0 m/s * sin(35°))) / 9.8 m/s^2]
Then, we can find the magnitude of the velocity at the time of impact using the Pythagorean theorem.
Velocity at impact = sqrt(Vx^2 + (Vertical component of velocity at impact)^2)
Hope this helps! Let me know if you have any further questions.
To solve this problem, we can use the equations of motion for projectile motion. The key variables we need to find are the horizontal range (x) and the velocity of the shell as it strikes the ground (v).
(a) To determine the horizontal range, we need to find the time of flight (t) first. We can use the vertical motion equation:
y = yo + voyt - 1/2gt^2
Where y = 0 (since the shell strikes the ground), yo = 30 m, voy = v*sin(theta), g = 9.8 m/s^2, and theta = 35°.
0 = 30 + (v*sin(35°))t - 1/2(9.8)t^2
Simplifying, we get:
4.9t^2 - (v*sin(35°))t - 30 = 0
This is a quadratic equation in terms of t. Solving this equation will give us two possible values for t. We can discard the negative value since time cannot be negative.
Using the quadratic formula, we get:
t = (-(v*sin(35°)) + sqrt((v*sin(35°))^2 - 4(4.9)(-30))) / (2(4.9))
Simplifying, we get:
t ≈ 5.57 s
Now that we have the time of flight, we can find the horizontal range using the equation:
x = voxt
Where x is the horizontal range, vox = v*cos(theta), and t = 5.57 s.
x = (v*cos(35°)) * 5.57
Simplifying, we get:
x ≈ 387.6 m
Therefore, the horizontal range of the shell is approximately 387.6 m.
(b) To determine the velocity of the shell as it strikes the ground, we can use the x and y components of velocity. The x component remains constant throughout the motion, while the y component changes due to acceleration.
The x component of velocity (vox) is given by:
vox = v*cos(theta)
Plugging in the values, we get:
vox = 75*cos(35°)
Simplifying, we get:
vox ≈ 61.3 m/s
The y component of velocity (voy) can be found using:
voy = v*sin(theta)
Plugging in the values, we get:
voy = 75*sin(35°)
Simplifying, we get:
voy ≈ 42.7 m/s
At the point of impact, the y component of velocity (vfy) is equal to -voy (since the velocity is downward).
Using the equation:
vfy = voy - gt
We can solve for vfy:
vfy = -voy + g*t
Plugging in the values, we get:
vfy = -42.7 + (9.8)(5.57)
Simplifying, we get:
vfy ≈ -45.7 m/s
The magnitude of the velocity vector v is given by the Pythagorean theorem:
v = sqrt(vox^2 + vfy^2)
Plugging in the values, we get:
v = sqrt((61.3)^2 + (-45.7)^2)
Using a calculator, we get:
v ≈ 76.3 m/s
Therefore, the velocity of the shell as it strikes the ground is approximately 76.3 m/s.
To solve this problem, we can break it down into a horizontal and vertical motion. Let's start with the horizontal motion.
(a) Determine the horizontal range of the shell:
In the absence of air resistance, the horizontal velocity of the shell remains constant throughout its flight. We can calculate it using the formula:
horizontal velocity (Vx) = muzzle speed (v) * cos(elevation angle)
Here, the muzzle speed of the shell is given as 75.0 m/s, and the elevation angle is given as 35°. So, the horizontal velocity is:
Vx = 75.0 m/s * cos(35°)
Using a calculator, we find that Vx is approximately 61.505 m/s.
Now, we can calculate the time of flight for the shell using the vertical motion.
The time of flight (T) for the shell can be found using the formula:
T = 2 * (vertical component of initial velocity) / acceleration due to gravity
The vertical component of the initial velocity (Vy) can be calculated using the formula:
Vy = muzzle speed (v) * sin(elevation angle)
So, the vertical component of the initial velocity is:
Vy = 75.0 m/s * sin(35°)
Using a calculator, we find that Vy is approximately 43.156 m/s.
The acceleration due to gravity (g) is approximately 9.8 m/s^2.
Substituting the values into the formula for the time of flight, we can calculate T:
T = 2 * 43.156 m/s / 9.8 m/s^2
Calculating T, we find that it is approximately 8.786 seconds.
Now, we can calculate the horizontal range (R) using the equation:
R = horizontal velocity (Vx) * time of flight (T)
Substituting the known values, we find:
R = 61.505 m/s * 8.786 s
Calculating R, we find that the horizontal range of the shell is approximately 540.19 meters.
(b) Determine the velocity of the shell as it strikes the ground:
The velocity of the shell as it strikes the ground will have both a horizontal and vertical component.
The final horizontal velocity (Vxf) remains constant and is equal to the initial horizontal velocity (Vx) calculated in part (a).
The final vertical velocity (Vyf) can be found using the equation:
Vyf = initial vertical velocity (Vy) - acceleration due to gravity * time of flight (T)
Substituting the known values, we have:
Vyf = 43.156 m/s - 9.8 m/s^2 * 8.786 s
Calculating Vyf, we find that it is approximately -43.156 m/s. Note the negative sign indicates that the shell is moving downward.
To calculate the magnitude of the velocity of the shell as it strikes the ground, we can use the Pythagorean theorem:
Velocity = sqrt(Vxf^2 + Vyf^2)
Substituting the values, we have:
Velocity = sqrt((61.505 m/s)^2 + (-43.156 m/s)^2)
Calculating the velocity, we find that it is approximately 75.0 m/s.