TG EAMCET 3 May 2025 Shift 2 Paper

Step 1: Calculate the speed after falling 20 mThe person starts from rest, so the initial speed is 0 . We use the formula:$v^2\;=u^2+2 g h$$v\;=\sqrt{2 \times 10 \times 20}$$v\;=\sqrt{400}=20\;\text{m/s}$This means after falling 20 meters, the person is moving at $20\;\text{m/s}$.Step 2: Find the time to fall the first 20 mUse the formula for distance with constant acceleration from rest:$s\;=u t+\;\frac{1}{2} g t^2$$20\;=0+\;\frac{1}{2} \cdot 10 \cdot t^2$$20\;=5 t^2$$t^2\;=4$$t\;=2\;\text{ seconds }\;$So, it takes 2 seconds to fall the first 20 meters.Step 3: Calculate the distance left after 20 mThe total height is 2000 meters. After falling 20 meters, the remaining distance is:Remaining distance $2000-20$$=1980\;\text{m}$Step 4: Find the time to travel the remaining 1980 m at constant speedNow, the person moves at $20\;\text{m/s}$ (from step 1). Time taken is:$t=\frac{1980}{20}=99\;\text{ seconds }\;$Step 5: Add both times to get the total time to reach the groundTotal time $=2+99=101$ seconds.