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.
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.
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