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.
Initial Velocity = 75 m/s
Elevation angle = 35°
Height of the cliff = 30 m
Range = (initial velocity) ^2 * sin(2θ)/g
Range = (75) ^2 * sin (2*35°) / 9.8 m/s
Range = 539.36 m
The horizontal range of the shell is approximately 539.36 m
(b) Determine the velocity of the shell as it strikes the ground.
Vertical Velocity = initial velocity * sin (θ)
Vertical velocity = 75 m/s * sin (35°)
Vertical Velocity = 42.8 m/s
The velocity of the shell is approximately 42.8 m/s
as it strikes the ground.
(c) Determine the time it takes for the shell to reach the ground.
We can use the equation of motion for vertical motion:
h = (v^2 * sin^2θ) / (2g)
where h is the initial vertical displacement (height of the cliff), v is the initial velocity (75 m/s), θ is the elevation angle (35°), and g is the acceleration due to gravity (9.8 m/s^2).
Plugging in the values, we have:
30 = (75^2 * sin^2(35°)) / (2 * 9.8)
Now we can solve for time by rearranging the equation:
t = √(2h / g)
t = √(2 * 30 / 9.8)
t ≈ √(6.12)
t ≈ 2.47 seconds
So, it takes approximately 2.47 seconds for the shell to reach the ground.
(d) Determine the horizontal and vertical components of velocity just before the shell hits the ground.
The horizontal component of velocity remains constant throughout the motion, so it is the same as the initial horizontal velocity:
Horizontal velocity = initial velocity * cos(θ)
Horizontal velocity = 75 m/s * cos(35°)
Horizontal velocity ≈ 61.4 m/s
The vertical component of velocity just before the shell hits the ground can be found using the equation:
v = u + at
where u is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time it takes for the shell to reach the ground (2.47 seconds).
Vertical velocity = initial vertical velocity + acceleration * time
Vertical velocity = 42.8 m/s + (-9.8 m/s^2) * 2.47 s
Vertical velocity ≈ 16.5 m/s (downward)
So, the horizontal component of velocity just before the shell hits the ground is approximately 61.4 m/s and the vertical component of velocity is approximately 16.5 m/s downward.
To solve part (a) - determining the horizontal range of the shell - we can use the following formula:
Range = (initial velocity) ^2 * sin(2θ) / g
In this case, the initial velocity is 75.0 m/s and the elevation angle is 35°. We also need the acceleration due to gravity, which is approximately 9.8 m/s².
Substituting the values into the formula, we get:
Range = (75.0)^2 * sin(2 * 35°) / 9.8
Simplifying further:
Range ≈ 539.36 m
Therefore, the horizontal range of the shell is approximately 539.36 meters.
To solve part (b) - determining the velocity of the shell as it strikes the ground - we can use the vertical velocity component:
Vertical Velocity = initial velocity * sin(θ)
Substituting the given values:
Vertical Velocity = 75.0 m/s * sin(35°)
Simplifying:
Vertical Velocity ≈ 42.8 m/s
Therefore, the velocity of the shell as it strikes the ground is approximately 42.8 m/s.