A ball is thrown straight up . Taking the drag force into account , does it take longer for the ball to travel to the top of its motion or for it to fall back..

Question

back down again ? 

 

 

Answer 1

Hi Essie!

 

We can take a look at this in two parts:  Newton's 2nd law (with some word arguments added in) and kinematics.

 

Newton's 2nd law

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The two significant forces in the vertical direction on this object are gravity and drag.  Gravity is always down, and drag is always opposite motion.  On the way up, then, gravity and drag must point in the same direction, and therefore reinforce each other:

  

F_net-up = -F_g-F_d = ma_u

 

On the way down, however, the drag force turns around and points upward (opposite the direction of motion.  Therefore, we have:

 

F_net-down = -F_g - F_d = ma_d

 

Because the drag now partly cancels out gravity, the resulting acceleration a_d must be smaller in magnitude than a_u.  On the way up, the object goes from some initial velocity v_o to 0.  On the way down, it goes from 0 initial velocity to some final velocity v_f at the ground, which is less than v_o, because drag has leeched away some energy.

 

Kinematics

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We can relate v_o (its velocity upon being thrown upward) and v_f (its velocity upon returning to the ground) using kinematics.  Since the magnitude of displacement up and displacement down is the same, let's use the kinematic equation:

 

v² = v_o²+2ad

 

For the upward motion, we have:

 

0 = v_o²+2(-a_u)d

 

where we know the final vertical velocity (at its highest point) is zero, and a_u is down (which I define to be negative) because both forces are down.  Rearranging this gives:

 

v_o² = 2a_ud

d = v_o²/2a_u

 

For the downward motion:

 

v_f² = 0 + 2(-a_d)(-d)

 

since the displacement is negative (object ends up lower than it starts), the object starts at rest from the top of its flight, and the acceleration (assuming terminal velocity is not reached) is still down as it falls.  From this, we get:

 

d = v_f²/2a_d

 

So, since the two displacements, up and down, must have equal magnitude, we can make the equation:

 

v_o²/2a_u = v_f²/2a_d 

 

v_o = v_f (√(a_u/a_d))

 

Since a_u>a_d in magnitude, as described above, this is another statement that the initial speed of throwing the object up (v_o) must be greater than its speed upon returning to the ground (v_f), if drag is present.

 

Then, we can pull out another kinematic equation:

 

d = (1/2)(v+v_o)t

 

For the upward leg, we have:

 

d = (1/2)(0+v_o)t_u = v_ot_u/2

 

And for the downward leg:

 

d = (1/2)(v_f + 0)t_d = v_ft_d/2

 

So we can equate (because the displacements are again equal in magnitude):

 

v_ot_u/2 = v_ft_d/2

 

t_d = (v_o/v_f)t_u

 

Since v_o > v_f, as described above, t_d (the time to come down) must be greater than t_u (the time to ascend).  Hence, it takes longer to come down.

 

I elided a few details here, mainly because we did not have an exact expression for the drag force in terms of velocity).  But if you have any questions, just let me know!

 

Answer 2

Without drag,  the time up equals the time down.   With drag, the average speed on the way up is greater than the average speed on the way down.  Since the distance covered is the same on the way up as it is on the way down, the trip down must take longer. 

A simple argument showing why the average speed on the way up is greater than the average speed on the way down is based on the energetics.   Without drag, speed at which the ball is launched is equal to its speed at the point where it has returned to the launch altitude.   This can be considered to be a consequence of the conservation of energy (potential plus kinetic).  However, when there is drag, mechanical energy is not conserved.  Some mechanical energy is lost to heat.  By this argument it is seen that the kinetic energy (and consequently the speed) at the point where the ball has returned to the launch altitude must be less than it was at launch.  A more detailed version of this argument shows that the speed on the way down must be less than the speed at the corresponding altitude on the way up for all corresponding points.  The one exception to this is at the top of the motion where the speed is zero.