New answers tagged

0

You have a net force $$F = F_G - F_D$$ $$ma = -mg + C_0 v^2$$ Dividing both sides by m, letting $C_1 = C_0 / m$, $$a = -g + C_1 v^2$$ $$\frac{dv}{dt} = -g + C_1 v^2$$ $$\frac{dv}{-g+C_1 v^2} = dt$$ Now to integrate both sides, we need to use the limits $[0,t]$ for $dt$, $[v_0,v]$ for $dv$ (where $v_0$ is our initial velocity. $$\int_{v_0}^{v} ...


2

If it's falling only then you have $F_d=+Cv^2$, where up is the positive direction. You said there is a gravitational force $F_g=-mg$. Write a Newton's 2nd Law equation, set $a=\frac{dv}{dt}$, rearrange, to get dv/g(v) = dt, (I'll let you find $g(v)$) and integrate away. The $v$ integral is not trivial. Look it up in an integral table, if your teacher will ...


1

The basketball and the medicine ball are both round, and though medicine balls can be a little larger than basketballs, the surface area and shape that each presents to air resistance is roughly equivalent. The major difference is weight. Basketballs weigh in the neighborhood of 20 ounces, but medicine balls can weigh up to 24 pounds. The experiment you ...


2

The problem is air. If something is lightweight compared to its surface area, like a feather, snowflake, dust mote, animal hair, it cannot fall as fast as a brick because air resistance slows it down. Even a brick or a bowling ball eventually reaches a maximum speed when falling in air. Any discussion of this has to assume (ideally) no air resistance, or at ...


0

$\newcommand{norm}[1]{\lVert #1 \rVert} \renewcommand{vec}[1]{\pmb{#1}}$Let's take a mathematical point of view. Let the mass of the objects be $m_i$ for $i=1,2$ and the length of the rod $l$. The total torque with respect to the center of mass (COM) is given by: $$\vec \tau \cdot \vec e_z = \vec e_z \cdot \left[ \vec r_1 \times \vec F_{g1} + \vec r_2 ...


0

Draw a graph with time along the horizontal and velocity up the vertical. Let's start with an object in motion at constant velocity. Its motion on the graph will be represented by a horizontal line at some distance from the y=0 axis. After some period of time, it will have covered a distance equal to velocity x time. That distance will be represented on ...


0

Start with acceleration, which we assume to be a constant $g$. Also assuming 'up' is the positive spacial direction i.e. $g$ is $(-)ve$: $$a(t) = -g = -9.81 \,ms^{-2}$$ Integrate once to get velocity: $$\dot{a} =v(t) = \int_0^t -g dt = -gt +v_0$$ Integrate again to get the distance: $$\ddot a = x(t) = \int_0^t(-gt + v_0)\, dt = -\frac{1}{2}gt^2 ...


1

You don't need calculus to understand this and I think you are right to be trying to gain a deeper understanding than just memorizing some formulas. During that first second the body accelerates - it starts with 0 velocity and gains linearly giving 9.8 at the end of the first second, so at that point, it hasn't been moving at 9.8m/s for a second, it has ...


0

Intuitively, the body spends some time at every velocity between 0 and 9.8 m/s during the first second. From the formula distance = speed * time, if we call that time interval dt and add up all the contributions (using integral calculus) the answer is 4.9 m.


1

Well let's say you weigh 220 lbs. Which translates over to 100 kg. The fall is 50 ft, so about 15.24 meters. The running thing that most people say is that it takes about 5000 Newtons of force to break a human bone, but we know that, this varies. We also know that it is not how hard you hit something that necessarily kills you, it's energy from the impact, ...


1

It's not the fall that kills you. Your chance of survival is 100% :) In regards to the suspected impact after the fall, you will need to expand on the parameters to your question. Divers survive 50ft drops routinely. The elderly are killed from 4 foot drops routinely. I'm sure this wasn't the answer you were looking for, but I think the best answer ...


2

Considering the way how $v^2=v_0 ^2-2g\Delta y$ is derived, It is derived through energy conservation, that is $$\frac{1}{2}mv^2-\frac{GMm}{r+\Delta y}=\frac{1}{2}mv_0^2-\frac{GMm}{r} \tag{1}$$ $$v^2=v_0^2-2\frac{GM}{r}\bigg(1-\frac{1}{1+\frac{\Delta y}{r}}\bigg)$$ Using Taylors series $$v^2=v_0^2-2GM\bigg(\frac{\Delta y}{r^2}-\frac{\Delta ...



Top 50 recent answers are included