Q124 / 180 · KCET 2011 · Physics
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 $y=8 t-5 t^{2} \mathrm{~m}$ and $x=6 t \mathrm{~m}$, where $t$ is in seconds. The velocity with which the projectile is projected is
- A$6 \mathrm{~ms}$
- B$8 \mathrm{~ms}^{1}$
- Correct answerC$10 \mathrm{~ms}^{-1}$
- D$14 \mathrm{~ms}^{-1}$
Solution
Given, $\quad y=8 t-5 t^{2} \quad \ldots$ (i)
We know,
$x=6 t$
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
$$
\begin{aligned}
u_{2} \sin \theta &=8 \\
u &=\sqrt{u_{1}^{2}+u_{2}^{2}} \\
u &=\sqrt{36+64} \\
u &=10 \mathrm{~ms}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.