An object is projected vertically up from ground at t = 0 in the presence of uniform gravity ( g = 10 m / s 2 ). It is observed to be at a given height at two instants of time t = 2 s and t = 4 s . Ma

Question

An object is projected vertically up from ground at $t = 0$ in the presence of uniform gravity ( $g = 10 m / s^{2}$ ). It is observed to be at a given height at two instants of time $t = 2 s$ and $t = 4 s$ . Match the column (All values are given in SI unit)

List - I	List - II
(I) Initial velocity of projection.	(P) 10
(II) Maximum height attained by particle.	(Q) 30
(III) Distance covered by particle in $t = 2 s$ to $t = 4 s$ .	(R) 45
(IV) Time taken to come back to initial point of projection.	(S) 6
	(T) None of these

TestHub · Accepted Answer

Equal heights at t = 2 s and t = 4 s put the peak at t = 3 s, so $u = 10 \times 3 = 30 m/s$ - I -> Q.

Maximum height: $H = \frac{u ^{2}}{2 g} = 45 m$ - II -> R.

From t = 2 s to 4 s the particle rises 5 m then falls 5 m: distance = 10 m - III -> P.

Time to return: $T = \frac{2 u}{g} = 6 s$ - IV -> S.

Correct option: (B)

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