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 the initial velocity of the shell into its horizontal and vertical components.
Given:
Initial vertical velocity (Vy) = 75.0 * sin(35°)
Initial horizontal velocity (Vx) = 75.0 * cos(35°)
Initial height (h) = 30.0 m
a) To determine the horizontal range of the shell, we need to find the time of flight (T). We can use the formula:
T = (2 * Vy) / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
T = (2 * (75.0 * sin(35°))) / 9.8
To find the horizontal range (R), we use the formula:
R = Vx * T
Substituting the values:
R = (75.0 * cos(35°)) * [(2 * (75.0 * sin(35°))) / 9.8]
Solve for R to get the horizontal range of the shell.
b) To determine the velocity of the shell as it strikes the ground, we need to find the final velocity (Vf) in the y-direction when the shell reaches the ground.
Using the formula:
Vf = Vy + gt
where t is the time of flight (T).
Vf = (75.0 * sin(35°)) + (9.8 * T)
Substitute the value of T we obtained in part a) to find the final y-direction velocity.
Finally, to find the velocity of the shell as it strikes the ground, we use the Pythagorean theorem:
V = √(Vx² + Vf²)
Substitute the values of Vx and Vf to calculate the final velocity of the shell as it strikes the ground.
To solve the problem, we'll break it down into two parts:
Part a) Determining the horizontal range of the shell:
Step 1: Split the initial velocity of the shell into horizontal and vertical components.
Given that the muzzle speed of the shell is 75.0 m/s and it is fired at an elevation of 35° above the horizontal, we can determine the horizontal and vertical components as follows:
Horizontal Component: V₀x = V₀ * cos(θ)
Vertical Component: V₀y = V₀ * sin(θ)
Where V₀ is the muzzle speed and θ is the angle of elevation.
Step 2: Determine the time it takes for the shell to reach the ground.
To find the time of flight, we can use the vertical component of the velocity since it determines the time of flight. First, we need to determine the time it takes for the shell to reach its maximum height. At this point, the vertical component of velocity becomes zero. Using the equation:
Vf = Vi + gt
Where Vf is the final velocity, Vi is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is time. Since the final velocity is zero, we can rearrange the equation to solve for time:
0 = V₀y - g * t
Simplifying the equation:
t = V₀y / g
Step 3: Calculate the time of flight.
The total time of flight is twice the time it takes for the shell to reach its maximum height. So, the time of flight is:
Time of Flight (t_flight) = 2 * t
Step 4: Calculate the horizontal range.
The horizontal range can be calculated using the horizontal component of velocity and the time of flight:
Horizontal Range (R) = V₀x * t_flight
Part b) Determining the velocity of the shell as it strikes the ground:
Step 1: Determine the vertical component of the final velocity.
The vertical component of the final velocity can be found using the equation:
Vf = Vi + gt
Where Vf is the final velocity, Vi is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time of flight.
Step 2: Calculate the horizontal component of the final velocity.
The horizontal component of the final velocity remains constant throughout the flight since no horizontal forces act on the shell. Therefore, the horizontal component of the final velocity is the same as the horizontal component of the initial velocity:
Vf_x = V₀x
Step 3: Calculate the final velocity.
The final velocity can be calculated using the horizontal and vertical components:
Vf = √(Vf_x² + Vf_y²)
Where Vf is the final velocity, Vf_x is the horizontal component of the final velocity, and Vf_y is the vertical component of the final velocity.
Now, let's plug in the values and calculate step by step:
Given:
Muzzle Speed (V₀) = 75.0 m/s
Angle of Elevation (θ) = 35°
Height of Cliff (h) = 30 m
Acceleration due to Gravity (g) = 9.8 m/s²
Step a):
Step 1:
Horizontal Component: V₀x = V₀ * cos(θ)
= 75.0 m/s * cos(35°)
≈ 61.44 m/s
Vertical Component: V₀y = V₀ * sin(θ)
= 75.0 m/s * sin(35°)
≈ 42.82 m/s
Step 2:
t = V₀y / g
= 42.82 m/s / 9.8 m/s²
≈ 4.37 s
Step 3:
Time of Flight (t_flight) = 2 * t
= 2 * 4.37 s
≈ 8.74 s
Step 4:
Horizontal Range (R) = V₀x * t_flight
= 61.44 m/s * 8.74 s
≈ 536.38 m
The horizontal range of the shell is approximately 536.38 m.
Step b):
Step 1:
Vf_y = Vi + gt
= V₀y + g * t_flight
= 42.82 m/s + 9.8 m/s² * 8.74 s
≈ 132.41 m/s
Step 2:
Vf_x = V₀x
= 61.44 m/s
Step 3:
Vf = √(Vf_x² + Vf_y²)
= √(61.44 m/s)² + (132.41 m/s)²
= √(3771.5136) + (17538.9481)
≈ √21310.4617
≈ 145.9 m/s
The velocity of the shell as it strikes the ground is approximately 145.9 m/s.
To solve these problems, we can break down the motion of the shell into its horizontal and vertical components.
First, let's determine the initial velocity of the shell in the horizontal and vertical directions:
Horizontal component:
The initial velocity in the horizontal direction (Vx) remains constant throughout the shell's flight because there is no horizontal acceleration. We can determine Vx using the initial muzzle velocity and the angle of elevation.
Vx = V * cos(θ)
V = 75.0 m/s (given)
θ = 35° (given)
Vx = 75.0 * cos(35°)
Vx ≈ 61.41 m/s
Vertical component:
The initial velocity in the vertical direction (Vy) can be determined using the initial muzzle velocity and the angle of elevation as well.
Vy = V * sin(θ)
V = 75.0 m/s (given)
θ = 35° (given)
Vy = 75.0 * sin(35°)
Vy ≈ 42.92 m/s
Now, let's solve each part of the problem:
a) Determining the horizontal range of the shell:
The horizontal range represents the total horizontal distance traveled by the shell before hitting the ground. We can use the horizontal component of the initial velocity (Vx) and time of flight (T) to calculate it.
T = 2 * (Vy / g)
g = 9.8 m/s² (acceleration due to gravity)
T = 2 * (42.92 m/s / 9.8 m/s²)
T ≈ 8.75 s
Range = Vx * T
Range = 61.41 m/s * 8.75 s
Range ≈ 536.91 m
Therefore, the horizontal range of the shell is approximately 536.91 meters.
b) Determining the velocity of the shell as it strikes the ground:
The final velocity (Vf) of the shell when it hits the ground can be calculated separately for the horizontal and vertical components.
Horizontal velocity remains constant, so Vfx = Vx = 61.41 m/s.
Vertical velocity can be determined using the equation:
Vfy = Vy - g * T
Vfy = 42.92 m/s - 9.8 m/s² * 8.75 s
Vfy ≈ -35.78 m/s
Note that we obtained a negative value for the vertical velocity (Vfy) because the shell is moving downward.
Finally, we can calculate the resultant final velocity (Vf) of the shell using the horizontal and vertical components:
Vf = √(Vfx² + Vfy²)
Vf = √((61.41 m/s)² + (-35.78 m/s)²)
Vf ≈ 70.10 m/s
Therefore, the velocity of the shell as it strikes the ground is approximately 70.10 m/s.