TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Given, speed of car at X,vx=10m∕s.
Acceleration due to gravity, g=10ms−2
Let, mass of car be m.Speed of car at X in downward direction (v1) be 0ms−1,
Speed of car at Z in downward direction be v2 and v be the speed of car at Z,
h1 and h2 are the position of X and Z from ground i.e., 100m and 60m, respectively.
Now, by using law of conservation of energy,
‌

1

2

mv22‌‌=mg(h1−h2)
⇒‌‌v2‌‌=√2g(h1−h2)
‌‌=√2×10(100−60)
‌‌=√2×10×40
‌‌=√800
‌‌=20√2ms−1
v‌‌=10

∧

i

+20√2

∧

j

‌‌|v|‌‌=√102+(20√2)2
‌‌=√100+400×2
‌‌=√900=30ms−1