- #1

rockinwhiz

- 28

- 6

- Homework Statement
- An experimenter throws a ball of mass ##m=1.0##kg vertically upwards with a velocity ##u=4.0m/s## from the top of a high tower. During the flight of the ball, modulus of the force of air resistance on the ball is given by the equation ##F=kv##, here ##k=0.41kg/s## and ##v## is the speed of the ball. The tower is so high that the ball achieves a constant speed before striking the ground. Find velocity of the ball, when its kinetic energy changes most rapidly. Acceleration due to gravity is ##g=10 m/s^2##

- Relevant Equations
- ##K_E= \frac 1 2 mv^2##

##F= ma##

##a= \frac {dv} {dt}##

Because, ##F=ma=kv##, therefore, ##a=kv/m##. Clearly, the net acceleration ##A=-(g+a)##.

Also, ##A=dv/dt=-(g+ \frac {kv} m )##, so cross multiplying and integrating LHS with respect to ##v## and RHS with respect to ##t## gives me:

$$ v= e^{ \frac {-tk} m } * (u + \frac {gm} k) - \frac {gm} k $$

Now, I'm not sure, but differentiating this with respect to time should gives the time when rate of change of ##K_E## is maximum.

This is where I need help. How do I proceed? And how does the terminal velocity come into play?

FYI, the given answer is: ##v \approx 12.2 m/s##

Thanks!

Also, ##A=dv/dt=-(g+ \frac {kv} m )##, so cross multiplying and integrating LHS with respect to ##v## and RHS with respect to ##t## gives me:

$$ v= e^{ \frac {-tk} m } * (u + \frac {gm} k) - \frac {gm} k $$

Now, I'm not sure, but differentiating this with respect to time should gives the time when rate of change of ##K_E## is maximum.

This is where I need help. How do I proceed? And how does the terminal velocity come into play?

FYI, the given answer is: ##v \approx 12.2 m/s##

Thanks!