Kinematics Questions concerning 2 particles thrown from a same point O but at different times

Question

The below is a question I have been given as part of a set of revision questions for an upcoming test. I understand the theory, and have been able to do the rest of the questions swiftly, but it’s this question that’s bothering me at the moment.

A stone is projected vertically upwards with a speed of 10ms-1. One second later, a second stone is projected from the same point with a speed of 8ms-1.

a) Find the velocity of the each stone when they meet.

So far, I have worked out that both stones meet 3.14m above the ground. This is at 1.7s for the first stone, and 0.7s for the second. How would I continue?

Answer 1

The first thrown stone $v_1=10 \frac{m}{s}$ can't be reached by the second stone $v_2=8 \frac{m}{s}$ because its speed is greater than the second stone and the second stone is thrown $\delta t=1 s$ later. So it gives ability to the first stone to reach the highest point A ($OA=H_1$) for the $t_1$ time. The 2 stones meeting point is B $OB=H_2$. Let's put equations of distance, time and speed for the 2 stones. Assume that accelaration of free fall equals $g = 9.8 \frac{m}{s^2}$ All variable names are case sensetive. We need to find $V_1, V_2$.

$ H_1 = v_1*t_1 - \frac{g*t_1^2}{2} (1)$

$ H_2 = v_2*t_3 - \frac{g*t_3^2}{2} (2)$

$ t_3=t_2-\delta t (3)$

$ t_2=t_1+T (4)$

$ V_1 = g*T (5)$

$ V_2 = v_2 - g*t_3 (6)$

$ AB = H_1-H_2 = \frac{g*T^2}{2} (7)$

$ 0 = v_1 - g*t_1 (8)$

The system of the equations (1) - (8) can be solved because we have 8 unknown variables and 8 equations. Let's solve the equations.

From (8) $t_1=\frac{v_1}{g} (8')$ so use (1') in (1) and we get $H_1=\frac{v_1^2}{2*g} (1')$

Use (1') and (2) in (7) $ \frac{v_1^2}{2*g}-v_2*t_3 + \frac{g*t_3^2}{2} = \frac{g*T^2}{2} (9)$

From (3) and (4) $t_3=t_1+T-\delta t (10)$

Let assign $R=t_1-\delta t =\frac{v_1}{g} -\delta t (10')$ using (8') so (10) can be rewritten such as $ t_3=R+T (10'')$

Use (10'') in (9) and simplify

$\frac{v_1^2}{2*g}-v_2*(R+T) + \frac{g*(R+T)^2}{2} = \frac{g*T^2}{2}$

$\frac{v_1^2}{2*g}-v_2*R-v_2*T + \frac{g}{2}*(R^2+2*R*T+T^2) = \frac{g*T^2}{2}$

$\frac{v_1^2}{2*g}-v_2*R-v_2*T + \frac{g}{2}*R^2+g*R*T+\frac{g*T^2}{2} - \frac{g*T^2}{2} = 0$

$\frac{v_1^2}{2*g}-v_2*R + \frac{g}{2}*R^2 = v_2*T-g*R*T$

$\frac{v_1^2}{2*g}-v_2*R + \frac{g}{2}*R^2 = (v_2-g*R)*T$

$T = \frac{\frac{v_1^2}{2*g}-v_2*R + \frac{g}{2}*R^2}{(v_2-g*R)} = 0.6334379905808477 s$

From (5) $V_1=6.207692307692308 \frac{m}{s}$

From (6) $V_2 = 1.5923076923076929 \frac{m}{s}$

Do you have questions? Do you find my errors? Let me know.