Question
A particle when thrown, moves such that it passes from same height at 2 and 10 seconds, then this height $$h$$ is :
A.
$$5\,g$$
B.
$$g$$
C.
$$8\,g$$
D.
$$10\,g$$
Answer :
$$10\,g$$
Solution :
As the time taken from $$D$$ to $$A = 2\,\sec .$$ and $$D \to A \to B \to C = 10\,\sec \,\left( {{\text{given}}} \right).$$
As ball goes from $$B \to C\left( {u = 0,t = 4\,\sec } \right)$$
$${v_c} = 0 + 4g.$$
As it moves from $$C$$ to $$D,$$ $$s = ut + \frac{1}{2}g{t^2}$$
$$s = 4g \times 2 + \frac{1}{2}g \times 4 = 10\,g.$$
As the time taken from $$D$$ to $$A = 2\,\sec .$$ and $$D \to A \to B \to C = 10\,\sec \,\left( {{\text{given}}} \right).$$
As ball goes from $$B \to C\left( {u = 0,t = 4\,\sec } \right)$$
$${v_c} = 0 + 4g.$$
As it moves from $$C$$ to $$D,$$ $$s = ut + \frac{1}{2}g{t^2}$$
$$s = 4g \times 2 + \frac{1}{2}g \times 4 = 10\,g.$$
