Question

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y=(8t-5t^2)$ metre and $x=6t$ metre, where $t$ is in second. The velocity of projection is

• A. 8 $m{s}^{-1}$
• B. 6 $m{s}^{-1}$
• C. 10 $m{s}^{-1}$
• D. $14m{s}^{-1}$

Answer 1

Answer: (C)

Given,
$y=8t-5t^2$...(i)
$x=6t$...(ii)
We know,
$x=(u\cos \theta )t$...(iii)
Compare with Eq. (ii), we get
$u_1\cos \theta =\frac{x}{t}=6$
and $y=(u\sin \theta )t-\frac{1}{2}g{t}^2$
Compare with Eq (i), we get
$u_2\sin \theta =8$
$\therefore u=\sqrt{u_1^2+u_2^2}$
$u=\sqrt{36+64}$
$u=10m{s}^{-1}$