Grade 11 physics: 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 first need to break down the given information into known values and determine what equations we can use to solve for the unknowns.
Known values:
- Initial velocity (muzzle speed) = 75.0 m/s
- Elevation angle = 35°
- Height of the cliff = 30 m
Unknowns:
(a) Horizontal range of the shell (displacement in the x-direction)
(b) Velocity of the shell as it strikes the ground (final velocity)
To determine the horizontal range of the shell, we can use the following equation:
Range = (Initial velocity)^2 * sin(2θ) / g
where θ is the elevation angle and g is the acceleration due to gravity (9.8 m/s^2).
(a) Using the given values, we have:
Range = (75.0 m/s)^2 * sin(2 * 35°) / (9.8 m/s^2)
Range ≈ 152.4 m
Therefore, the horizontal range of the shell is approximately 152.4 m.
To determine the velocity of the shell as it strikes the ground, we need to find the vertical component of the velocity at that point. Using the equation for vertical velocity in projectile motion, we have:
Vertical velocity = Initial velocity * sin(θ)
(b) Using the given values, we have:
Vertical velocity = 75.0 m/s * sin(35°)
Vertical velocity ≈ 42.8 m/s
Now, since projectile motion is symmetrical, the horizontal component of the velocity remains constant throughout the motion. Therefore, the velocity of the shell as it strikes the ground will have the same horizontal component as the initial velocity.
Thus, the velocity of the shell as it strikes the ground is approximately 75.0 m/s.
To solve this problem, we can break it down into two components - the horizontal component and the vertical component.
(a) Determining the horizontal range of the shell:
1. Start by finding the horizontal component of the initial velocity (Vx).
Vx = V_initial * cos(theta)
Vx = 75.0 m/s * cos(35°)
Vx = 61.33 m/s (rounded to 2 decimal places)
2. Use the horizontal component of the initial velocity and the time of flight (tf) to calculate the horizontal range (R).
tf = 2 * (V_initial * sin(theta)) / g
tf = 2 * (75.0 m/s * sin(35°)) / 9.8 m/s^2
tf ≈ 9.80 s (rounded to 2 decimal places)
3. Finally, calculate the horizontal range using the formula:
R = Vx * tf
R = 61.33 m/s * 9.80 s
R ≈ 600.07 m (rounded to 2 decimal places)
Therefore, the horizontal range of the shell is approximately 600.07 m.
(b) Determining the velocity of the shell as it strikes the ground:
1. Start by finding the vertical component of the final velocity (Vy).
Vy = V_initial * sin(theta)
Vy = 75.0 m/s * sin(35°)
Vy = 42.90 m/s (rounded to 2 decimal places)
2. Since the shell is fired from a height of 30 m, we need to account for the vertical distance it has fallen. Use the equation:
h = (1/2) * g * t^2
where h is the height (30 m) and t is the time of flight (9.80 s, rounded from earlier).
Rearrange the equation to solve for the time of flight (t):
t^2 = 2h / g
t^2 = 2 * 30 m / 9.8 m/s^2
t ≈ 2.0364 s (rounded to 4 decimal places)
3. Calculate the vertical component of the final velocity (Vfy) using the equation:
Vfy = Vy + (g * t)
Vfy = 42.90 m/s + (9.8 m/s^2 * 2.0364 s)
Vfy ≈ 62.09 m/s (rounded to 2 decimal places)
4. Finally, calculate the magnitude of the final velocity using the horizontal and vertical components:
V_final = sqrt(Vx^2 + Vfy^2)
V_final ≈ sqrt((61.33 m/s)^2 + (62.09 m/s)^2)
V_final ≈ 87.18 m/s (rounded to 2 decimal places)
Therefore, the velocity of the shell as it strikes the ground is approximately 87.18 m/s.
To solve this projectile motion problem, we can break it down into two components: horizontal and vertical motion.
(a) Determining the horizontal range of the shell:
The horizontal motion of the shell is not affected by gravity since there is no force acting in the horizontal direction. Therefore, we can use the formula for the horizontal distance:
Range = Velocity_horizontal * Time
To find the horizontal component of the velocity (Vx), we can use trigonometry:
Vx = Velocity * cos(angle)
Given:
- Velocity (V) = 75.0 m/s
- Angle (θ) = 35°
Calculating Vx:
Vx = 75.0 m/s * cos(35°)
Vx ≈ 75.0 m/s * 0.819
Vx ≈ 61.425 m/s
Now, let's calculate the time (t) it takes for the shell to hit the ground:
In vertical motion, the shell will experience free fall, so we can use the formula:
y = y0 + Vyt - 0.5 * g * t^2
where:
- y = 0 m (since the shell hits the ground)
- y0 = 30 m (the height of the cliff)
- Vy = Velocity_vertical
- g = acceleration due to gravity (approximated as 9.8 m/s^2)
Rearranging the equation and solving for t:
Yielding equation: 30 m + Vyt - 0.5 * 9.8 m/s^2 * t^2 = 0
Using the quadratic formula: t = (-b ± sqrt(b^2 - 4ac)) / 2a
a = -4.9 m/s^2 (0.5 * g)
b = Vy
c = y0
t = (-Vy ± sqrt(Vy^2 - 4(-4.9)(30))) / 2 * (-4.9)
t = (-Vy ± sqrt(Vy^2 + 588)) / 9.8
Since we are interested in the positive value of t when the shell hits the ground, we can ignore the negative value in this case.
Substituting Vy = V * sin(θ):
t = (-V * sin(θ) + sqrt((V * sin(θ))^2 + 588)) / 9.8
Now we can substitute the known values and solve for t:
t = (-75.0 m/s * sin(35°) + sqrt((75.0 m/s * sin(35°))^2 + 588)) / 9.8
t ≈ (-43.4 + sqrt(1844 + 588)) / 9.8
t ≈ (-43.4 + sqrt(2432)) / 9.8
t ≈ (-43.4 + 49.32) / 9.8
t ≈ 5.92 s
Now we can calculate the horizontal range:
Range = Vx * t
Range ≈ 61.425 m/s * 5.92 s
Range ≈ 363.618 m
Therefore, the horizontal range of the shell is approximately 363.618 meters.
(b) Determining the velocity of the shell as it strikes the ground:
To find the velocity of the shell as it strikes the ground, we can use the formula:
V = sqrt(Vx^2 + Vy^2)
Substituting the known values:
V = sqrt((61.425 m/s)^2 + (75.0 m/s * sin(35°))^2)
V = sqrt(3771.184 + 2022.704)
V ≈ sqrt(5793.888)
V ≈ 76.06 m/s
Therefore, the velocity of the shell as it strikes the ground is approximately 76.06 m/s.