Let's write the vertical displacement for the first and second ball:
y1=v01t−21gt2,y2=v02(t−Δt)−21g(t−Δt)2.The two balls will meeting when y1=y2:
v01t−21gt2=v02(t−Δt)−21g(t−Δt)2,v01t−21gt2=v02t−v02Δt−21gt2+gtΔt−21gΔt2,(v01−v02−gΔt)t=−v02Δt−21gΔt2,t=(−v02Δt−21gΔt2)(v01−v02−gΔt),t=(−25 ms−1⋅1 s−21⋅9.8 ms−2⋅(1 s)2)(10 ms−1−25 ms−1−9.8 ms−2⋅1 s)=0.83 s.As we find the time at what the two balls were met, we can find the height:
y=10 sm⋅0.83 s−21⋅9.8 s2m⋅(0.83 s)2=4.92 m.Answer:
y=4.92 m.
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