Question

A projectile is fired horizontally from a gun that is 45.0 m above flat ground, it emerges from the gun with a speed of 500 m/s. (take, g = $-10m/s^2$)

Column - I Column - II

(i) Time of flight of Projectile in s (x) 3

(ii) Horizontal distance covered by projectile in m (y) 30

(iii) Magnitude of vertical component of velocity in m/s as it strikes ground(z) 1500

• A. (i) - (x); (ii) - (z); (iii) - (y)
• B. (i) - (y); (ii) - (x); (iii) - (z)
• C. (i) - (x); (ii) - (y); (iii) - (z)
• D. (i) - (z); (ii) - (x); (iii) - (y)

Answer 1

The correct option is A

(i) - (x); (ii) - (z); (iii) - (y)

So the time for which the ball was in the air is the time that the ball took to fall 45 m in the y - direction.

Well that's fair to say,

So let's apply equation of motion.

$s$ = $ut+\frac{1}{2}a{t}^2$

$s_y$ = -45, $a_y$ = $-10m/s^2$

$u_y$ = $0$ $\Rightarrow$ $-{45}^9$ = $-\frac{1}{2}\times {10}^2\times t^2$

$t$ = $3sec$

so the ball was in air for 3 sec

time of flight = 3 sec

now we need to find the horizontal distance covered by the ball.

So let's see what we know about the motion in x

$u_x$ = 500 m/s

$a_x$ = 0 No acceleration in x-direction

The ball was in air for 3 sec

$\Rightarrow$ t = 3

$s_x$ = $u_{x}t$

= $500\times 3$

$s_x$ = $1500m$

Next part of the equation asks the vertical compound of velocity when the ball hits the ground.

So I need to focus on the motion in y-direction only and should not care about x-direction as they both are independent of each other

So I know

$u_y$ = $0$

$a_y$ = $-10m/s^2$

$s_y$ = $-45m$

$t$ = $3sec$

what I need is $V_y$-equation of motion

$v$ = $u+at$

Simple

$v_y$ = $-10\times 3$

$v_y$ = $-30m/s$

So the vertical copmponent of the ball when it hits the ground is $30m/s$ in the negative y-direction.