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)

10 $ms^{ - 1}$

Answer 2

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