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 $ ms^{ - 1} $
• B. 6 $ ms^{ - 1} $
• C. 10 $ ms^{ - 1} $
• D. $14\,ms^{-1}$

Answer 1

Answer: (C)

Given,
$y=8 t-5 t^{2}$...(i)
$x=6 t$...(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
$u_{2} \sin \theta =8$
$\therefore u =\sqrt{u_{1}^{2}+u_{2}^{2}}$
$u =\sqrt{36+64}$
$u =10\, ms ^{-1}$