Question-209099

Question Number 209099 by Spillover last updated on 01/Jul/24

Answered by mr W last updated on 02/Jul/24

say particle 2 starts time T later than particle 1. particle 2 at time t: h=ut−((gt^2 )/2) particle 1 at time t+T: h=u(t+T)−((g(t+T)^2 )/2) when both particles meet at height h, u(t+T)−((g(t+T)^2 )/2)=ut−((gt^2 )/2) ⇒t=((2u−gT)/(2g)) h=ut−((gt^2 )/2)=((2u−gT)/(2g))×(u−(g/2)×((2u−gT)/(2g))) =((4u^2 −g^2 T^2 )/(8g)) ✓

$${say}\:{particle}\:\mathrm{2}\:{starts}\:{time}\:{T}\:\:{later} \\ $$$${than}\:{particle}\:\mathrm{1}.\: \\ $$$$ \\ $$$${particle}\:\mathrm{2}\:{at}\:{time}\:{t}: \\ $$$${h}={ut}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${particle}\:\mathrm{1}\:{at}\:{time}\:{t}+{T}: \\ $$$${h}={u}\left({t}+{T}\right)−\frac{{g}\left({t}+{T}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${when}\:{both}\:{particles}\:{meet}\:{at}\:{height}\:{h}, \\ $$$${u}\left({t}+{T}\right)−\frac{{g}\left({t}+{T}\right)^{\mathrm{2}} }{\mathrm{2}}={ut}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{t}=\frac{\mathrm{2}{u}−{gT}}{\mathrm{2}{g}} \\ $$$${h}={ut}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{2}{u}−{gT}}{\mathrm{2}{g}}×\left({u}−\frac{{g}}{\mathrm{2}}×\frac{\mathrm{2}{u}−{gT}}{\mathrm{2}{g}}\right) \\ $$$$\:\:\:=\frac{\mathrm{4}{u}^{\mathrm{2}} −{g}^{\mathrm{2}} {T}^{\mathrm{2}} }{\mathrm{8}{g}}\:\:\checkmark \\ $$

Commented by Spillover last updated on 02/Jul/24