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?
80 m
heightfall=1/2 g t^2
To find the vertical distance the cannonball will have fallen after 4 seconds, we can use the formula for the distance traveled under constant acceleration:
d = (1/2) * g * t^2
Where:
d = vertical distance
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
In this case, we'll assume the cannonball is launched horizontally, so the initial vertical velocity is 0. Therefore, the vertical distance fallen is computed independently of the horizontal velocity.
Plugging in the values:
d = (1/2) * 9.8 * (4^2)
d = 4.9 * 16
d = 78.4 meters
Therefore, neglecting air resistance, the cannonball will have fallen approximately 78.4 meters vertically after 4 seconds.
To find out how far the cannonball will have fallen vertically after 4 seconds, we need to use the formula for vertical displacement under freefall:
d = (1/2) * g * t^2
where:
d = vertical displacement
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (in seconds)
Since the cannonball is launched horizontally, its initial vertical velocity is zero. Therefore, we can omit the equation involving initial vertical velocity.
Now, let's plug in the given values into the equation:
d = (1/2) * 9.8 * (4^2)
d = (1/2) * 9.8 * (16)
d = 78.4 meters
Therefore, the cannonball will have fallen approximately 78.4 meters vertically after 4 seconds.