Question

A projectile is fired horizontally with a velocity $98\text{m/s}$ from the top of a hill $490\text{m}$ high. Find the coordinates of the projectile at time $t=10\text{sec}$.
Take origin at the top of hill and vertically downwards as positive $y$-direction, positive $x$-direction along the direction of horizontal projection.$[g=9.8{\text{m/s}}^2]$

• A. $(980\text{m},-245\text{m})$
• B. $(980\text{m},-122.5\text{m})$
• C. $(490\text{m},-490\text{m})$
• D. $(980\text{m},490\text{m})$

Answer 1

The correct option is D $(980\text{m},490\text{m})$


Considering origin at $\text{O}$
Initial velocity in $x$-direction:
$u_x=+98\text{m/s}$
Acceleration in $x-$ direction, $a_x=0{\text{m/s}}^2$
Initial velocity in $y$- direction, $u_y=0\text{m/s}$
Acceleration in $y-$ direction, $a_y=+g$

at $t=10\text{s}$ displacement in $x$ direction:
$S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$S_x=98\times 10+\frac{1}{2}\times 0\times (10{)}^2$

$\therefore [S_x=980\text{m}]$

Displacement in $y$- direction:
$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$S_y=0\times 10+\frac{1}{2}\times (+9.8)\times (10{)}^2$
$=\frac{1}{2}\times (+9.8)\times 100$

$\therefore [S_y=490\text{m}]$

Coordinates of the projectile after $t=10\text{s}$ is ($S_x,S_y$)

$\Rightarrow (S_x,S_y)=(980,490)\text{m}$

But height of the hill is $490\text{m}$
So, projectile at $t=10\text{sec}$ touches the ground at point $A$ as shown in figure.