Motion along a straight line problem 4

Free fall

A circus juggler throws a ball upward with an initial velocity of 20.0 m/s.

1)How long does he have to wait before it comes back down to the height from which he threw it?

2)What is the maximum height above the level of throwing that the ball will reach? 

 

We denote quantities in this problem as follows:

 

– initial velocity of ball,

– maximum height (above the level of throwing) reached,

– time from the moment of throwing to the moment of return to the same level.

 

Schematic drawing of the situation is in

Fig. M1-P4.1

 

It is convenient to divide the problem into two parts:

a)the ball moving up to the moment it stops (and then starts falling down),

b)the ball falling down from the maximum height reached.

 

 

The magnitude of h₀ from the Fig. M1-P4.1 does not have any influence on the result, we can neglect it.

 

This is motion along a straight line with initial velocity v₀ and negative acceleration g=9.81m/s². Negative means having direction opposite to velocity. 

The equation required for this situation is Eq. M1.30, which originally reads

 

x = x₀ + v₀ t+(a t² / 2)    

 

and for this problem can be rewritten as

 

h_max=v₀t_r – g t_r² / 2   (1)

 

because we do not include in the solution the initial height, and the acceleration is negative, as explained above.

denotes the time of rising of the ball.

 

The velocity of a ball during moving up will obey Eq. M1.28, which has the form

 

v(t) = v₀ + at

 

but for our problem it must be written as

 

v(t) = v₀ – g t   (2)

 

The final velocity at time t_r is equal zero. The ball simply stops for a moment before it starts falling down. Substituting into Eq. 2 v(t_r) = 0 we get

 

0 = v₀ – g t_r  

 

 and from this we can calculate the time of rising of the ball

 

t_r = v₀ / g     (3)

 

Substituting this time into Eq. 1 we will find the maximum height reached by the ball

 

h_max=v₀ (v₀ / g) – g (v₀ / g)² / 2

 

and, after rearranging

 

h_max= v₀² / g –(v₀ g / g)² / 2 = v₀² / g  – v₀² / (2 g)

 

h_max = (1/2) v₀² / g      (4)

 

This is the answer to question 2) from our problem. We leave the substituting of numbers to the Reader. We will only check units

[h_max ]=(m / s)² /( m / s²) =( m² / s²) *( s² / m) = m

 

which is correct.

 

Now question 1) – the total time of “flight” of the ball. We already calculated the time of rising to the maximum height – Eq. 3. Now we need the time of falling from that height to the starting level. The distance to travel is the same – h_max, and now it is a free fall with initial velocity zero, so the time to travel this distance is given by Eq. M1.33

 

t_f = (2 h_max / g )^1/2   (5)

t_f – time of free falling from distance h_max.

 

After substituting h_max from Eq. 4 into Eq. 5 we will have

 

t_f = v₀ / g    (6)

 

The total time of “flight” is

t = t_r + t_f = 2v₀ / g      (7)

 

Answer to question 1) is given by Eq. 7 and to question 2) by Eq. 4. It is worth to notice that the time of rising (Eq. 3) is equal to the time of falling (Eq. 6). Knowledge of this can be used to simplify any other problems involving free fall motion.

Remember, free fall motion is not only when the object is decreasing its height, but also when it is moving vertically upwards under the influence of gravity. We can think about such object as falling down with velocity being the sum of free fall velocity and that of initial velocity.

Analogically if we throw the object vertically down from some elevation, it is also a case of free fall.  

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