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:
$A= 22.14285714$
$B=-7.142857143$
$t_{max} = 0.8081443652$ (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?
Recall that $$v_y(t) = \frac{dy}{dt}\Rightarrow y(t)-y_0=\int_{t_0}^{t} v_y \,dt.$$