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.
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.
1 answer