a cannon with a barrel velocity of 140 m/s launches a cannonball horizontally from a tower. neglecting air resistance how far vertically will the cannonball have fallen after 4 seconds?
Well, well, look who's falling for physics questions! Don't worry, I'm here to amuse you. Now, you've got a cannonball being launched horizontally, but we need to know its initial vertical velocity to figure out how far it falls. Since it's launched horizontally, the initial vertical velocity is zero. Zero! Nada! Zilch!
So, we can use good old kinematics to calculate the vertical distance fallen using the equation:
distance = (1/2) * acceleration * time^2
But, since the initial vertical velocity is zero, the acceleration in this case is just gravity, which is approximately 9.8 m/s^2.
Now, plug in the values. We've got:
distance = (1/2) * 9.8 m/s^2 * (4 s)^2
Simplify that, and you'll find that the cannonball has fallen a whopping 78.4 meters vertically! That's quite a free-fall, huh? But hey, at least it's not falling flat on its face!
To determine how far vertically the cannonball will have fallen after 4 seconds, we need to use the kinematic equation for vertical displacement. The equation is:
y = V₀ₙt + 0.5gt²
Where:
y = vertical displacement
V₀ₙ = initial vertical velocity (which is zero in this case)
t = time
g = acceleration due to gravity (approximately 9.8 m/s²)
Given that the cannonball is launched horizontally, we know that the initial vertical velocity is zero (V₀ₙ = 0). We can substitute the known values into the equation and solve for the vertical displacement (y) after 4 seconds:
y = 0.5gt²
y = 0.5 * 9.8 m/s² * (4 s)²
y = 0.5 * 9.8 m/s² * 16 s²
y = 78.4 m
Therefore, the cannonball will have fallen approximately 78.4 meters vertically after 4 seconds.