Answer to Question #155422 in Mechanics | Relativity for Jude Judith

Question #155422

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


1
Expert's answer
2021-01-15T05:14:07-0500

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


y1=v01t12gt2,y_1=v_{01}t-\dfrac{1}{2}gt^2,y2=v02(tΔt)12g(tΔt)2.y_2=v_{02}(t-\Delta t)-\dfrac{1}{2}g(t-\Delta t)^2.

The two balls will meeting when y1=y2y_1=y_2:


v01t12gt2=v02(tΔt)12g(tΔt)2,v_{01}t-\dfrac{1}{2}gt^2=v_{02}(t-\Delta t)-\dfrac{1}{2}g(t-\Delta t)^2,v01t12gt2=v02tv02Δt12gt2+gtΔt12gΔt2,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,(v01v02gΔt)t=v02Δt12gΔt2,(v_{01}-v_{02}-g\Delta t)t=-v_{02}\Delta t-\dfrac{1}{2}g\Delta t^2,t=(v01v02gΔt)(v02Δt12gΔt2),t=\dfrac{(v_{01}-v_{02}-g\Delta t)}{(-v_{02}\Delta t-\dfrac{1}{2}g\Delta t^2)},t=(10 ms125 ms19.8 ms21 s)(25 ms11 s129.8 ms2(1 s)2)=0.83 s.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 ms0.83 s129.8 ms2(0.83 s)2=4.92 m.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.y=4.92\ m.


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