TG EAMCET 3 May 2025 Shift 1 Paper

Step 1: Use the conservation of energyWhen an object moves in Earth's gravity, its total energy (kinetic plus potential) stays the same if no energy is lost.Step 2: Write the energy equationThe equation is: $\;\frac{1}{2} m V_i^2-\;\frac{G M m}{R}=\;\frac{1}{2} m V_f^2+0$Here, $V_i$ is the speed at the start, and $V_f$ is the speed far away from Earth (where gravity is almost zero).Step 3: Substitute the initial speedThe body is thrown with speed $V_i=\sqrt{5} V_e$, where $V_e$ is escape velocity.Step 4: Plug in values and simplify$\;\;\frac{1}{2} m (\sqrt{5} V_e)^2-\;\frac{g R^2 m}{R}=\;\frac{1}{2} m V_f^2$$\;\Rightarrow\;\frac{5}{2} m V_e^2-g R m=\;\frac{1}{2} m V_f^2$$\;\Rightarrow\;\frac{5}{2} V_e^2-g R=\;\frac{1}{2} V_f^2$Step 5: Express $g R$ in terms of $V_e$Remember, $V_e=\sqrt{2 g R}$. Square both sides to get $V_e^2=2 g R$.So, $g R=\;\frac{V_e^2}{2}$.Step 6: Substitute and solve for $V_f$Plug $g R=\;\frac{V_e^2}{2}$ into the previous equation:$\;\;\frac{5}{2} V_e^2-\;\frac{V_e^2}{2}=\;\frac{1}{2} V_f^2$$\;(5-1)\;\frac{V_e^2}{2}=\;\frac{1}{2} V_f^2$$\;4\;\frac{V_e^2}{2}=\;\frac{1}{2} V_f^2$Multiply both sides by 2 :$4 V_e^2=V_f^2$Take square root:$V_f=2 V_e$