Question

A particle is projected with some angle from the surface of the planet. The motion of the particle is described by the equation; $x=t, y=t-t^2$. Then match the following columns:

Column I (quantity)		Column II (magnitude only)
A	Velocity of projection	1	1
B	Acceleration	2	$\sqrt{2}$
C	Time of flight	3	2
D	Maximum height attained	4	$\frac{1}{4}$

• A. (A) $\rightarrow$(4) ; (B) $\rightarrow $(1) ; C $\rightarrow$(2) ; (D) $\rightarrow$ (2)
• B. (A) $\rightarrow$(2) ; (B) $\rightarrow$(3) ; C $\rightarrow$(1); (D) $\rightarrow$ (2)
• C. (A) $\rightarrow$ (2); (B) $\rightarrow$ (3); C$ \rightarrow$ (1); (D) $\rightarrow$ (4)
• D. ( A )$ \rightarrow$ (3) ;( B )$ \rightarrow$ (4) ; C$ \rightarrow$ (3) ;( D ) $\rightarrow$ (2)

Answer 1

Answer: (C)

$( A ) \rightarrow(2) ;( B ) \rightarrow(3) ; C \rightarrow(1) ; (D) \rightarrow(4)$
$U_x=\frac{d x}{d t}=1 \text { and } U_y=\frac{d x}{d t}=1-2 t$
$\therefore U_{t=0}=\sqrt{u_x^2+y_y^2}=\sqrt{1^2+1^2}=\sqrt{2} m / s$
$a_x=\frac{d^2 x}{d t^2}=0 ; d_y=\frac{d^2 y}{d t^2}=-2$
For time of flight,
$y=0 \text { or } 0=t-t^2 \therefore t=1 s \text {. }$
For maximum height, $t=\frac{1}{2} s$
$\therefore H=t-t^2=\frac{1}{2}-\left(\frac{1}{2}\right)^2=\frac{1}{4} m$