Question

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (assuming flat surface) with no surrounding atmosphere are given by $x=6t$ and $y=8t-5t^2$, where $x$ and $y$ are in metre and time $t$ is in second. Find the velocity in $m{s}^{-1}$ with which the body is projected. Take acceleration due to gravity $g=10{\text{m/s}}^2$

• A. 10.000
• B. 10
• C. 10.0
• D. 10.00

Answer 1

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

Let the acceleration due to gravity on that planet be $g^{\prime }$
$u_x=u\cos \theta$
$u_y=u\sin \theta$
$x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$x=u_{x}t$ (since horizontal acceleration is zero)
$y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
Putting $a_y=-g^{\prime }$
$y=u_{y}t-\frac{1}{2}{g}^{\prime }{t}^2$
Comparing with the given data
$x=6t$ and $y=8t-5t^2$

$u_x=6$ and $u_y=8$
$\therefore u^2=6^2+8^2$
$⟹u=10m/s$