The second identical cannon ball was dropped
from the tower with zero initial speed.
Which cannon ball will hit the ground first?
Please, do not ignore air resistance.
2007-07-17 04:49:13 · 11 answers · asked by Alexander 6 in Science & Mathematics ➔ Physics
The vertical gravitational acceleration is the same for both balls, so we only need to examine the deacceleration due to air resistance. Suppose both balls have the same vertical velocity, while the shot ball has a horizontal velocity. Then the deacceleration would be proportional to some function of the ball's velocity. If the function is linear, then the vertical deacceleration due to air resistance would be the same. However, if the function is any power of the velocity higher than 1, then the vertical deacceleration would be greater for the shot ball. So, the dropped ball would reach the ground first, since at high velocities the function becomes non-linear.
2007-07-17 05:58:20 · answer #1 · answered by Scythian1950 7 · 1⤊ 0⤋
Theoretically they will it the ground together; the vertical component of air resistance is the same if it is a function of
vertical velocity. But air resistance is not that simple. It is non-linear, has vortices, and modeling problems. The high
horizontal air resistance may carry over to the vertical and so
the second ball may hit first.
I don't know.
Why don't you go up a 175m tower with a cannon
and two balls and let us know the outcome?
2007-07-17 05:18:00 · answer #2 · answered by ? 5 · 1⤊ 0⤋
If air resistance is NOT ignored, it is impossible to predict which will land first. This would depend entirely on the spin imparted on the cannonball by the cannon. Like a baseball, a spinning cannonball will react with the air around it. If the spin is in the right direction, the cannon can create an area of high pressure beneath it and low pressure above that will keep it aloft slightly longer than the dropped cannon ball. If the spin is the other way, the fired cannonball will sink faster than its dropped counterpart.
2007-07-17 05:07:24 · answer #3 · answered by dansinger61 6 · 1⤊ 1⤋
The time to reach ground depend only upon the motion of the balls in vertical direction.
Both balls experience the same air resistance in the vertical direction.
The horizontal speed may reduce due to air resistance, but the time to fall vertically is not affected by this reduction of horizontal speed.
Considering the air resistance, since the balls are identical both will reach the ground in the same time since the friction is the same for both.
2007-07-17 05:54:36 · answer #4 · answered by Pearlsawme 7 · 0⤊ 1⤋
Ignoring air resistance: a. Do this "time of free-fall" formula to find the time spent in free-fall; time = square-root of (2h/g) [double height divided by acceleration due to gravity (the acceleration due to Earths gravity is a constant 9.81m/s² towards its center.)] = 4.28 s b. Once you answer a. (time of free-fall), then simply multiply that by the speed at which the ball is travelling to get the distance. = 1,500 m c. Now, I'm not sure about this, because, they could be asking for the vertical velocity or the horizontal velocity. In the direction of the balls parabolic path the velocity is greater than either the vertical or horizontal velocity. The horizontal velocity is still 350 m/s since there's no air resistance and gravity only accelerates in the vertical -not horizontal- axis. Thus, vx = 350 m/s The vertical velocity can be found with this formula; vy = square-root of (2gh) [double gravity times height] or by multiplying the "time of free-fall" formula by g (acceleration due to gravity). vy = -42.0 m/s (negative because it's travelling down and velocity is a vector quantity, meaning a direction is needed.) To find the velocity in the direction of the balls parabolic path, an instant before it hits the ground, we do a trigonometry formula; net v = vy / sine(inverse tangent(vy/vx)) net v = 353 m/s at 6.8° below horizontal (we get the angle from the "inverse tangent(vy/vx)" part of the formula). I hope i didn't stuff anything up.
2016-04-01 08:38:52 · answer #5 · answered by Anonymous · 0⤊ 0⤋
The dropped ball would land first.
The tower is 175m (574 feet). Gravity's pull is 32 ft.sec. Terminal velocity is approx.120mph (176 ft sec). Therefore the dropped ball would require a little more than 4 seconds to hit the gound.
The fired ball would reach terminal velocity and beyond instantly. The speed/distance as you have asked cannot be equated accurately, as we would need to know the energy of the cannon explosion (amount of gunpowder vs weight of ball)
2007-07-17 05:24:05 · answer #6 · answered by Anonymous · 0⤊ 1⤋
They hit the ground at the same time,.
All objects experience free falling near the earth's surface experience the same force of gravity, (and due to their shape and density cannon balls experience minimal friction) so they accelerate downwards at the same rate and hit the ground at the same time as horizontal and vertical motion under gravity are two independent quantities
2007-07-17 05:02:30 · answer #7 · answered by Mandél M 3 · 0⤊ 2⤋
assuming you meant do not ignore air resistance like you typed (unlike the people above me) the one fired horizontally because it would be going faster and thus have more air resistance
2007-07-17 05:09:06 · answer #8 · answered by Anonymous · 0⤊ 1⤋
He said NOT to ignore air resistance!!!
What is wrong with you people!!!
Are you deaf???
2007-07-17 05:51:58 · answer #9 · answered by Anonymous · 3⤊ 1⤋
I guess you meant : please ignore air resistance...
in these conditions, both balls will touch ground at the same time, reason being that they both are subject to the same gravitationnal attraction.
2007-07-17 05:01:22 · answer #10 · answered by Wilfried V 2 · 0⤊ 3⤋