A ball is thrown upwards from a point on the ground, with an initial velocity $v_0$. the ball is affected by earth's gravity, and force of fraction with air that depends on the velocity of the ball. fraction force is $Ff = -αmv$, while $m$ is the mass of the ball, $v$ is the momentary velocity, and $α$ is positive constant. $y$ is the vertical axis with the the positive direction upwards.

$g = 10 m/sec^{-2}$

$v_0 = 15 m sec^{-1}$

$α = 1.4 sec^{-1} $

the euqation of movement is : $\dot{v}_y + \alpha v_y + g = 0$

the general solution of the movement equation is : $v_{y}(t)=Ae^{-\alpha t}+B$

after solving the first sub questions I found that:

$A= 22.14285714$


$t_{max} = 0.8081443652$ (the time that is required for the ball to reach maximum height)

(untill here there was no need to use "$Ff = -αmv$")

the last part of the question asks: what is the maximal height that the ball can reach?


1 Answer 1


Recall that $$v_y(t) = \frac{dy}{dt}\Rightarrow y(t)-y_0=\int_{t_0}^{t} v_y \,dt.$$

  • $\begingroup$ $\int_{v_0}^{v(t)} v_y \,dt.$ = $\frac{-1}{α}e^{-αt} + Bt$ right? so by substituting $t_{max} = 0.8081443652$ I get a negative number, and I know its not the right answer. what could be wrong ? $\endgroup$ Commented Nov 8, 2014 at 15:14
  • $\begingroup$ I'm sorry, I committed a mistake. Already corrected it. $\endgroup$ Commented Nov 8, 2014 at 16:11
  • 1
    $\begingroup$ However the integral you calculated isn't correct. $$\int_{t_0}^t Ae^{-\alpha t}+B dt = \left[-\frac{A}{\alpha}e^{-\alpha t} + Bt \right]^t_{t_0}.$$ $\endgroup$ Commented Nov 8, 2014 at 16:12
  • $\begingroup$ then what is the right integral ? must be + C ? $\endgroup$ Commented Nov 8, 2014 at 16:12

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