Mathematical modelling that involves projectile motion I was asked to solve a mathematical differential equation to find the time taken by an object to reach the highest point and the time taken by the object to fall from its highest point to ground. I found that the time taken for the object to descend from highest point is longer than the time it being thrown up. Anyone knows why tis happens? what conclusion shall i make?
iwas asked to show differential equation as $m(dv/dt)=-\alpha v-mg$,
after that calculate the time to ascend and descend ,the mass of the object,$m$ is $0.1$ kg , initial velocity $u=5 \text{m/s}$ and $α=1.0$ kg/s and $g=10$ m/s$^2$.
 A: The solution to that differential equation (where $v(t)=\frac{dy}{dt}$) is:
$$ y(t)=-\frac{gm}{\alpha}t-\frac{m}{\alpha}(u+\frac{gm}{\alpha})e^{-\frac{\alpha}{m}t}+y(0),$$ where $y(0)$ is the initial position which we'll say is zero. A plot of this solution is attached. If you find the maxima of $y(t)$, i.e. set the derivative to zero and solve for $t$, you find that the ascending time is $0.179$ seconds. If you then determine when the projectile will hit the ground, i.e. when $y(t)=0$ you'll see this takes another $0.419$ seconds after it reaches its maximum height. In other words, it takes the projectile $0.179$ seconds to ascend and $0.419$ seconds to descend. Qualitatively this is seen in the figure below. It likely takes longer to descend because of the drag term--if you set $\alpha=0$ in the original differential equation (in other words saying there's no drag) the solution reduces to a parabolic function where the time to ascend equals the time to descend.
Hope this helps!
Cheers,
Paul Safier

A: When you throw it up all the force is at the beginning of the journey so you have to throw it hard to overcome gravity and all the drag it will experience, hence to start with the ball will be moving very fast and hence will get there quickly.
On the way down gravity and drag are operating in opposite directions, if it is a long drop the ball will approach its terminal velocity and come down at a steady rate.
If you have ever played badminton think of the trajectory of the shuttlecock after a lob shot.
A: So we are talking about physics. Very well :-) A generic solution (supposing the initial velocity $v_0$ is vertical and pointing up is
$$
y=v_0t -\frac{1}{2}gt^2
$$
the object will reach the maximum height when the vertical velocity (derivative of the previous equation) is zero so we have
$$
t_{max}=\frac{v_0}{g}
$$
so the maximum height is 
$$
y_{max}=\frac{v_0^2}{2g}
$$
Now it will be again on the ground when $y=0$ and therefore from our equation
$$
t_{end}=\frac{2v_0}{g}
$$
and you will notice that is exactly two times the time needed to reach the maximum. So the two times are the same and equal to $v_0/g$. Hope this help you. If you need help in solving the differential equation let me know and I will try to point you in the right direction.
A: The problem is solved symbolically in P.Glaister Times of Flight.
Glaister says that even with the assumption that $V<U$, (since energy will not be conserved) I have yet to deduce directly that $T_u<T_d\,$ or $\,T_u>T_d$ .
($U$ is the speed of projection, $V$ is the speed of return, $T_u$ is the "time up", $T_d$ is the "time down")
The well known result $T_d>T_u$ is shown using Taylor's theorem.
It can be useful to read also this pdf by A. Jobbings with its geometric approach, whatever the resistive force.
