Grade 11 physics: 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'll break it down into two parts: the horizontal motion and the vertical motion.
a) Determining the horizontal range:
The horizontal range is the horizontal distance covered by the projectile before it hits the ground.
We can use the equation for horizontal range: R = Vx * t
where R is the range, Vx is the horizontal component of the initial velocity, and t is the time of flight.
First, we need to find Vx, the horizontal component of the initial velocity:
Vx = V * cos(θ)
where V is the initial velocity and θ is the launch angle.
Given that V = 75.0 m/s and θ = 35°:
Vx = 75.0 m/s * cos(35°)
Now we need to find t, the time of flight:
t = 2 * t_max
where t_max is the time it takes for the shell to reach its maximum height.
We can use the equation for the time it takes to reach maximum height: t_max = Vy / g
where Vy is the vertical component of the initial velocity and g is the acceleration due to gravity (9.8 m/s^2).
To find Vy, the vertical component of the initial velocity:
Vy = V * sin(θ)
Given V = 75.0 m/s and θ = 35°:
Vy = 75.0 m/s * sin(35°)
Now we can solve for t_max and t:
t_max = Vy / g
t = 2 * t_max
Finally, we can calculate the horizontal range R:
R = Vx * t
b) Determining the velocity of the shell as it strikes the ground:
The velocity of the shell as it strikes the ground is the resultant velocity.
We can use the equation for the resultant velocity: V = sqrt(Vx^2 + Vy^2)
where Vx is the horizontal component of the initial velocity and Vy is the vertical component of the initial velocity.
Now we can find the horizontal range (part a) and the velocity of the shell as it strikes the ground (part b) using the given values and equations.
To solve this problem, we will need to break down the projectile motion into its horizontal and vertical components.
(a) Determining the horizontal range:
The horizontal component of the velocity remains constant throughout the motion, so we can calculate the horizontal range by multiplying the horizontal component of the velocity by the time of flight.
First, let's find the horizontal component of the velocity (Vx):
Vx = V * cos(θ)
where V is the muzzle velocity and θ is the angle of elevation.
Vx = 75.0 m/s * cos(35°) = 61.39 m/s (rounded to two decimal places)
Next, let's find the time of flight (t):
The time it takes for the shell to reach the ground can be calculated using the equation:
t = 2 * Vy / g
where Vy is the vertical component of the velocity and g is the acceleration due to gravity (9.8 m/s^2).
Since the shell is fired at an angle above the horizontal, the initial vertical component of the velocity (Vy) can be calculated using:
Vy = V * sin(θ)
Vy = 75.0 m/s * sin(35°) = 42.72 m/s (rounded to two decimal places)
Now we can substitute Vy into the equation for time of flight:
t = 2 * 42.72 m/s / 9.8 m/s^2 ≈ 8.71 s (rounded to two decimal places)
Finally, we can calculate the horizontal range (R):
R = Vx * t = 61.39 m/s * 8.71 s ≈ 535.14 m (rounded to two decimal places)
Therefore, the horizontal range of the shell is approximately 535.14 meters.
(b) Determining the velocity of the shell as it strikes the ground:
To determine the velocity of the shell as it hits the ground, we need to find the final velocity, which is the vector sum of the horizontal and vertical components of the velocity.
For the vertical component, vfy (vertical final velocity), we can use the equation:
vfy = Vy - g * t
vfy = 42.72 m/s - 9.8 m/s^2 * 8.71 s ≈ -31.84 m/s (rounded to two decimal places)
(Note: The negative sign indicates the direction downward.)
Now, to find the magnitude of the final velocity (vf), we can use the Pythagorean theorem:
vf = sqrt(Vx^2 + vfy^2)
vf = sqrt((61.39 m/s)^2 + (-31.84 m/s)^2 ) ≈ 69.18 m/s (rounded to two decimal places)
Therefore, the velocity of the shell as it strikes the ground is approximately 69.18 m/s.
To solve this problem, we can break it down into two components: the horizontal motion and the vertical motion of the shell.
(a) Determining the horizontal range:
The horizontal range refers to the horizontal distance covered by the shell before it hits the ground. To find this distance, we need to determine the time it takes for the shell to hit the ground.
First, let's find the time of flight. We can use the vertical motion equation:
h = v₀y * t - (1/2) * g * t²
Where:
h = vertical displacement = -30 m (negative because it's below the starting point)
v₀y = initial vertical velocity = v₀ * sin(θ)
g = acceleration due to gravity = 9.8 m/s²
t = time of flight (what we want to find)
Plugging in the known values:
-30 = (75 * sin(35°)) * t - (1/2) * 9.8 * t²
Rearranging the equation:
(1/2) * 9.8 * t² - (75 * sin(35°)) * t - 30 = 0
Now we have a quadratic equation. We can solve it using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Where:
a = (1/2) * 9.8 = 4.9
b = -75 * sin(35°)
c = -30
Solving for t, we will get two values (positive and negative). We discard the negative value since time cannot be negative.
Now that we have the time of flight (t), we can calculate the horizontal range using the horizontal motion equation:
x = v₀x * t
Where:
x = horizontal range (what we want to find)
v₀x = initial horizontal velocity = v₀ * cos(θ)
t = time of flight
Plugging in the known values:
x = (75 * cos(35°)) * t
Substituting t with the positive value we obtained earlier, we can calculate the horizontal range.
(b) Determining the velocity at impact:
The velocity at which the shell strikes the ground includes both the horizontal and vertical components of velocity.
The horizontal component remains constant throughout the flight, which is given by:
v_x = v₀x
The vertical component can be found using the vertical motion equation:
v_y = v₀y - g * t
Where:
v_y = final vertical velocity (what we want to find)
v₀y = initial vertical velocity = v₀ * sin(θ)
g = acceleration due to gravity = 9.8 m/s²
t = time of flight
Substituting the known values, we can calculate the final vertical velocity.
Finally, we can find the magnitude of the velocity at impact by calculating the resultant velocity using the Pythagorean theorem:
v = √(v_x² + v_y²)
Substituting the calculated values, we can find the velocity at impact.