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

Answer 1

Answer: (C)

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