The scenario that I'm having is such that a ball of radius $15mm$ is thrown from a location point $\vec{p}=(2, 5, 2)$ in a direction of $\vec{d}=(3, 0, 4)$. The initial velocity is $30m/s$. There were wind velocity at $15m/s$ in the direction of $\vec{v}=(0, 1, 0)$. The mass of the ball is $2kg$ and the gravity is $9.81m/s^2$.

I'm trying to calculate the time for the ball to hit the plane at $z=0$.

I started by making the throw direction as an unit vector: $\hat{\vec{d}} = \frac{\begin{bmatrix} 3\\ 0\\ 4 \end{bmatrix}}{\left | \begin{bmatrix} 3\\ 0\\ 4 \end{bmatrix} \right |}=\begin{bmatrix} \frac{3}{5}\\ 0\\ \frac{4}{5} \end{bmatrix}$

This is where the ball will reach at time $t$: $\vec{r(t)} = \vec{p}+t \cdot \hat{\vec{d}}$

$r_z = 2+t \cdot \frac{4}{5}$

Since equation of plane is $z=0$, if $z<15$, ball hits the plane:

$r_z < 15 \Rightarrow 2+t \cdot \frac{4}{5} < 15 \Rightarrow t\geq16.25$

This would imply that when $t\geq16.25$ seconds, the ball hits the plane. But I realise this doesn't consider the mass, gravity and wind resistance! So this value is probably invalid from start.

I recalled that $F=ma$, so $F= 2kg \cdot 9.81m/s^2$. But $F$ is just a scalar in this case and I don't know how I can use it any where.

How can I calculate the time $t$ for when the ball hits the plane with the consideration of the mass, gravity and wind resistance?


2 Answers 2


The correct equation for position at time $t$ : $\vec{r(t)} = \vec{p} + \vec{v}.t + {1\over2}\vec{a}.t^2$

Where $\vec{v}$ accounts for both initial speed and wind speed, thus: $\vec{v} = 30.\hat{\vec{d}} + 15.\hat{\vec{v}} = (18,15,24)$

And $\vec{a} = (0,0,-9.81)$ is gravity acceleration assuming it's pointed toward -z axis.

It hits the ground when z equates the radius, so: $2 + 24t - {9.81\over2}t^2 = 0.015$

Solving for the above equation give you the time. Mass doesn't play a role because we know the acceleration from the start, also the same for wind velocity as long as it has no z component.

Edit: We could calculate the drag force for a sphere: $F_d = {1\over2}c_d \rho v^2 A$

In which $c_d$ is a geometrical constant that is equal to 0.47 for sphere, $\rho$ is the density of the medium (air in this case), $v$ speed relative to air, $A$ is the cross section area ($\pi r^2$).

From that you could calculate acceleration due to drag (that's where mass comes in) and you need to integrate manually or (numerically) (look at JJ Fleck solution for this part) acceleration equation to get position equation as the above equation for $\vec{r(t)}$ is valid only for the case of constant acceleration.

  • $\begingroup$ The question asks for a full treatment including wind resistance, which this answer does not provide. $\endgroup$ Apr 27, 2013 at 23:51
  • $\begingroup$ As I said before windspeed just add up to the initial speed. While I've already done that it has no effect on motion along z. $\endgroup$
    – Azad
    Apr 28, 2013 at 5:08
  • $\begingroup$ And for air resistance we need to have air resistance coefficient of the ball even for an approximate answer. $\endgroup$
    – Azad
    Apr 28, 2013 at 5:09
  • $\begingroup$ "While I've already done that it has no effect on motion along z" That is incorrect. And yes, there is no general solution. $\endgroup$ Apr 28, 2013 at 5:59
  • $\begingroup$ That is incorrect Why? It has no z component. My major is physics and I'm completely sure about such rudimentary questions. $\endgroup$
    – Azad
    Apr 28, 2013 at 6:15

Let's assume wind resistance force is linear in speed with respect to the air (correct at usual "low" speed) with resistance coefficient called $\alpha$ (we will then be able to get an analytic answer) so that it is written $$\vec{f}=-\alpha\left(\vec{v}-\vec{v}_{\text{wind}}\right)$$

Note that this term exists even if wind speed is zero.

Applying Newton second law (and not forgetting the weight) $$ m \frac{d\vec{v}}{dt} = -\alpha\left(\vec{v}-\vec{v}_{\text{wind}}\right) + m\vec{g} $$

Projecting on the $z$-axis knowing that $\vec{v}\cdot\vec{e_z}=v_z$, $\vec{v}_{\text{wind}}\cdot\vec{e_z}=0$ and $\vec{g}\cdot\vec{e_z}=-g$, we get

$$ m\frac{dv_z}{dt} = -\alpha\,v_z - mg $$

that is

$$ v_z(t) = -g\tau + (v_{z0}+g\tau)\,e^{-t/\tau} $$

where $\tau = \frac{m}{\alpha}$ and $v_{z0}$ is the initial $z$ speed. Integrating once again gives

$$ z(t) = -g\tau\,t + (v_{z0}+g\tau)\,\tau\,(1-e^{-t/\tau}) + z_0 $$

with $z_0$ the initial $z$ position. You just have to solve this (numerically or graphically) to get the time $t_1$ where $z(t_1)=0$.


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