Does the mass of a ball matter when determining the time it takes to fall? Why or why not?
4 years ago
Answered By Leonardo F
It depends if we consider or not the presence of drag force, or air resistance. Is we disregard air resistance, as many Physics 20 problems do, then the mass of the object will NOT matter. In free fall, one of the equations that governs the motion is Torricelli's equation:
$vf^2=vi^2+2ad$vƒ2=vi2+2ad
In which vf if the final velocity, vi is the initial velocity, a is the acceleration (in this case, the gravitational acceleration, g), and d is the distance.
We can also apply the position equation for a uniform accelerated motion:
In which t will be the time of the fall. Notice that the mass does not participate in any equation.
If we consider air resistance, or drag force, then the physics gets more complicated, because the time of the fall WILL DEPEND on the mass of the object, and also the geometry of the body.
In which Cd is the drag coefficient, density is the density of the air surrounding the object, and A is the normal area of the object. Since the time of the fall will depend on the velocity of the object and the velocity will change, because the object is falling, the calculation is done with iterations, estimating the velocity, calculating the drag coefficient and then the time of the fall, until it converges to a value.
4 years ago
Answered By Jackie C
A:
No, as when the ball falls, its potential energy is transfered into kinetic energy, i.e.,
mgh=mv^2/2, thus v=sqrt(2gh), on the other hand,
g=v/t, therefore, t=v/g=sqrt(2gh)/g=sqrt(2h/g), where g is gravatation aqcceleration
There is nothing to do with its mass m.
4 years ago
Answered By Leonardo F
I made a mistake in the equation for the time of the fall with drag force. The right equation is:
4 years ago
Answered By Leonardo F
It depends if we consider or not the presence of drag force, or air resistance. Is we disregard air resistance, as many Physics 20 problems do, then the mass of the object will NOT matter. In free fall, one of the equations that governs the motion is Torricelli's equation:
$vf^2=vi^2+2ad$vƒ 2=vi2+2ad
In which vf if the final velocity, vi is the initial velocity, a is the acceleration (in this case, the gravitational acceleration, g), and d is the distance.
We can also apply the position equation for a uniform accelerated motion:
$d=\left(vi\right)\left(t\right)+\frac{1}{2}\left(a\right)\left(t\right)^2$d=(vi)(t)+12 (a)(t)2
In which t will be the time of the fall. Notice that the mass does not participate in any equation.
If we consider air resistance, or drag force, then the physics gets more complicated, because the time of the fall WILL DEPEND on the mass of the object, and also the geometry of the body.
The time of the fall can be calculated by:
$t=\frac{tanh^{-1}\left(\frac{v}{\sqrt{\frac{v}{m.g.k}}}\right)}{\sqrt{\frac{g.k}{m}}}$t=tanh−1(v√vm.g.k )√g.km
In which m is the mass of the object, g is the gravitational acceleration, v is the velocity of the body and k is defined by:
$k=\frac{1}{2}\left(Cd\right)\left(density\right)\left(A\right)$k=12 (Cd)(density)(A)
In which Cd is the drag coefficient, density is the density of the air surrounding the object, and A is the normal area of the object. Since the time of the fall will depend on the velocity of the object and the velocity will change, because the object is falling, the calculation is done with iterations, estimating the velocity, calculating the drag coefficient and then the time of the fall, until it converges to a value.
4 years ago
Answered By Jackie C
A:
No, as when the ball falls, its potential energy is transfered into kinetic energy, i.e.,
mgh=mv^2/2, thus v=sqrt(2gh), on the other hand,
g=v/t, therefore, t=v/g=sqrt(2gh)/g=sqrt(2h/g), where g is gravatation aqcceleration
There is nothing to do with its mass m.
4 years ago
Answered By Leonardo F
I made a mistake in the equation for the time of the fall with drag force. The right equation is:
$t=\frac{tanh^{-1}\left(\frac{v}{\sqrt{\frac{m.g}{k}}}\right)}{\sqrt{\frac{g.k}{m}}}$t=tanh−1(v√m.gk )√g.km