Answer in Mechanics | Relativity for Prosper Mawuli #108572
March 12th, 2023
Let time taken by first particle be “t.”
If second particle is projected with velocity“=3frac{m}{sec}”
For first particle, “d=frac{1}{2}gt^2”
For second particle, “d=3(t-3)+frac{1}{2}g(t-3)^2”
Equating both the equations, we get
“frac{1}{2}gt^2=3(t-3)+frac{1}{2}g(t-3)^2”
“t=frac{4}{3}sec”
Putting “t=frac{4}{3}sec” in any of the equations, we get “d=frac{80}{9}m”
