Question

A particle is projected at point $A$ with initial velocity $5\text{m/s}$ at an angle $\theta ={37}^{\circ }$ with the vertical $y-$axis. A strong horizontal wind gives the particle a constant horizontal acceleration of $6{\text{m/s}}^2$ in the $x-$direction. If the particle strikes the ground directly under its released position, then choose the correct option(s).

$[g=10{\text{m/s}}^2,\sin {37}^{\circ }=\frac{3}{5},\cos {37}^{\circ }=\frac{4}{5}]$

• A. Height $h$ of point $A$ is $9\text{m}$
• B. The time taken by the particle to reach point $B$ is $1\text{s}$
• C. The time taken by the particle to reach point $B$ is $2\text{s}$
• D. Height $h$ of point $A$ is $6\text{m}$

Answer 1

Answer: (A), (B)

The time taken by the particle to reach point $B$ is $1\text{s}$

Initial velocity of particle:
$u_x=-5\sin \theta =-5\sin {37}^{\circ }=-3\text{m/s}$
$u_y=5\cos \theta =5\cos {37}^{\circ }=4\text{m/s}$

Also,
$a_x=6{\text{m/s}}^2$
$a_y=10{\text{m/s}}^2$

Let us suppose the time taken by the particle to reach the bottom is $t$. Then,

$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$\Rightarrow h=4t+\frac{1}{2}\times 10\times t^2$

$\Rightarrow h=4t+5t^2...\left(i\right)$

Further, the horizontal displacement will be zero when the particle hits the ground.

So,

$S_x=0$

$\Rightarrow 0=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$\Rightarrow 0=-3t+\frac{1}{2}\times 6\times t^2$

$\Rightarrow 0=-t+t^2$

$\Rightarrow t=0\text{(not possible)}$ or $t=1\text{s}$

Therefore, the time taken by the particle to reach point $B$ is $1\text{s}$.

From $\left(i\right)$, height of point $A$,

$h=4\times 1+5\times 1^2=9\text{m}$

Hence, height $h$ of point $A$ is $9\text{m}$.