Question

A ball is thrown vertically upward with a speed of 10m/s.1sec later from the same position ball b is thrown vertically upward at the same path with the speed of 25m/s.what is the height of ball b when it meets ball a

Answer 1

Let's write the vertical displacement for the first and second ball:

$$y_1=v_{01}t-\dfrac{1}{2}gt^2,$$ $$y_2=v_{02}(t-\Delta t)-\dfrac{1}{2}g(t-\Delta t)^2.$$

The two balls will meeting when $y_1=y_2$:

$$v_{01}t-\dfrac{1}{2}gt^2=v_{02}(t-\Delta t)-\dfrac{1}{2}g(t-\Delta t)^2,$$ $$v_{01}t-\dfrac{1}{2}gt^2=v_{02}t-v_{02}\Delta t-\dfrac{1}{2}gt^2+gt\Delta t-\dfrac{1}{2}g\Delta t^2,$$ $$(v_{01}-v_{02}-g\Delta t)t=-v_{02}\Delta t-\dfrac{1}{2}g\Delta t^2,$$ $$t=\dfrac{(v_{01}-v_{02}-g\Delta t)}{(-v_{02}\Delta t-\dfrac{1}{2}g\Delta t^2)},$$ $$t=\dfrac{(10\ ms^{-1}-25\ ms^{-1}-9.8\ ms^{-2}\cdot 1\ s)}{(-25\ ms^{-1}\cdot 1\ s-\dfrac{1}{2}\cdot 9.8\ ms^{-2}\cdot(1\ s)^2)}=0.83\ s.$$

As we find the time at what the two balls were met, we can find the height:

$$y=10\ \dfrac{m}{s}\cdot 0.83\ s-\dfrac{1}{2}\cdot 9.8\ \dfrac{m}{s^2}\cdot (0.83\ s)^2=4.92\ m.$$

Answer:

$y=4.92\ m.$