Vertical launch and free fall

Preliminary considerations

We are faced with a problem that combines a free fall motion (first stone) with a downward vertical launch (second stone). Both are uniformly accelerated rectilinear motions (u.a.r.m.) taking place in the vertical axis (y-axis). So, we will only have to use the following two equations:

We place the system of reference at the bottom of the cliff, as shown in the following figure:

The following table shows the data given in the statement:

	Stone I	Stone II
Final velocity v	?	?
Initial velocity v0	0	?
Accelerationn a	g = -10 m/s2	g = -10 m/s2
Time t	t1	t2 = t1 - 1
Final position y	0	0
Initial positiony0	80	80

 

We start by substituting the data in the equation of position of the u.a.r.m. for the first stone and we get the time that it takes the first stone to reach the ground.

$$0=80+\frac{1}{2}(-10){t_1}^2 \Rightarrow t_1=\sqrt{\frac{80·2}{10}}=4 \text{sg}$$

For the calculation of the velocity, we simply apply the equation of velocity of the u.a.r.m. using the data in the table, and considering the value of t₁ = 4 s:

$$v_{f1}=(-10)·4 =\hspace{0.17em}-40 \text{m/s}$$

The negative velocity indicates the downward direction of the stone. With this, we have responded the second question.

Since the second stone arrives to the ground same time as the first one but is released a second later, the time that it is in the air is a second less that the first one, therefore:

$$t_2=t_1-1 =\hspace{0.17em}4-1=3 \text{sg}$$

With this, we have responded the third question.

To answer the first question, we simply apply the equation of position in the u.a.r.m. for the second stone, considering the value of t₂ = 3 sg:

$$0=80+v_{02}·3+\frac{1}{2}(-10)(3{)}^2\Rightarrow v_{02}=\frac{45-80}{3}=-11.6 \text{m/s}$$

With this we answer the first question.