The shell’s trajectory curves downward because of gravity, so at the time t = 0.217 s
the shell is below the straight line by some
vertical distance ∆h.
Find this distance ∆h in the absence of
air resistance. The acceleration of gravity is
9.8 m/s
2
.
Answer in units of m
Well, let's clown around and find out the vertical distance ∆h! To do that, we'll need to use some projectile motion and a dash of physics humor.
We know that the downward acceleration due to gravity is 9.8 m/s², and we can assume that it remains constant throughout. So, let's calculate how far the object would fall in the given time.
Using the equation for displacement with constant acceleration:
∆h = (1/2) * g * t²
Plugging in the values:
∆h = (0.5) * (9.8 m/s²) * (0.217 s)²
∆h ≈ 0.234 m
So, the vertical distance below the straight line (without considering air resistance) is approximately 0.234 meters.
Remember, this answer is purely hypothetical and doesn't account for real-world factors like air resistance. Keep your head up and enjoy the physics circus!
To find the vertical distance ∆h in the absence of air resistance, we can use the kinematic equation:
Δh = (1/2)gt^2
Where:
Δh is the vertical distance
g is the acceleration due to gravity (9.8 m/s^2)
t is the time (0.217 s)
Let's substitute the given values into the formula to find ∆h:
Δh = (1/2)(9.8)(0.217)^2
Δh = 1/2 * 9.8 * 0.04689
Δh = 0.2297 m
Therefore, the vertical distance ∆h in the absence of air resistance is approximately 0.2297 meters.
To find the vertical distance (∆h) below the straight line at time t = 0.217 s, we can use the equations of motion under constant acceleration. In this case, the vertical acceleration is due to gravity and is equal to 9.8 m/s^2.
The equation that relates the vertical displacement (∆h), initial velocity (u), time (t), and acceleration (a) is:
∆h = ut + 0.5at^2
Since the shell is initially at rest when it is released, its initial velocity (u) is 0 m/s. Therefore, the equation simplifies to:
∆h = 0.5at^2
Now we can substitute the values given in the problem. The acceleration due to gravity (a) is -9.8 m/s^2 (negative here because it acts downward), and the time (t) is 0.217 s. Hence, the equation becomes:
∆h = 0.5 * (-9.8) * (0.217)^2
Now we can calculate:
∆h = 0.5 * (-9.8) * (0.047089) ≈ -0.226 m
Therefore, the vertical distance ∆h below the straight line at time t = 0.217 s, in the absence of air resistance, is approximately 0.226 meters (m) downwards.
∆h = (g/2)t^2
Do the numbers