Question

In the projectile motion shown is figure, given $t_{AB}=2s$ then which of the folowing is correct : ($g=10m{s}^{-2}$)

• A. particle is at point B at $3s$
• B. maximum height of projectile is $20m$
• C. initial vertical component of velocity is $20m{s}^{-1}$
• D. horizontal component of velocity is $20m{s}^{-1}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A particle is at point B at $3s$
B maximum height of projectile is $20m$
C initial vertical component of velocity is $20m{s}^{-1}$
D horizontal component of velocity is $20m{s}^{-1}$

Let $u$ and $v$ be the horizontal and vertical components of the initial velocity vector. Now, $u=\frac{AB}{t_{AB}}$, where $AB=40m$. Therefore $u=20m/s$ and as a result, $t_{OA}=1s$. From the kinematic equations of motions, $s=vt-\frac{1}{2}g{t}^2$, where $s=15m$, we get $v=20m/s$. And from $0^2=v^2-2gh$, we get, maximum height of projectile($h$) = $20m$. $t_{OB}=t_{OA}+t_{AB}=3s$.