Question

A particle is projected from the ground with some initial velocity making an angle of $45^{\circ}$ with the horizontal. If it reaches a height of $7.5\, m$ above the ground while it travels a horizontal distance of $10\, m$ from the point of projection, then the initial speed of particle is
(Assume $g =10 \,m / s ^{2}$ )

• A. $10\, m / s$
• B. $20\, m / s$
• C. $15 \,m / s$
• D. $25\, m / s$

Answer 1

Answer: (B)

The given situation is as shown in figure,

Let the particle reaches point $P$ in time $t$.
Then, horizontal distance, $x=u \cos \theta t$
$ \Rightarrow 10=u \cos 45^{\circ} \times t$
$\Rightarrow t=\frac{10 \sqrt{2}}{u} \ldots$ (i)
and vertical distance,
$y=u \sin \theta t-\frac{1}{2} g t^{2} $
$\Rightarrow 7.5=u \sin 45^{\circ} \times t-\frac{1}{2} \times 10 \times t^{2} \ldots$ (ii)
Substituting value of $t$ from Eq. (i) into Eq. (ii), we get
$\Rightarrow 7.5=\frac{u}{\sqrt{2}} \times \frac{10 \sqrt{2}}{u}-\frac{1}{2} \times 10 \times\left(\frac{10 \sqrt{2}}{u}\right)^{2} $
$\Rightarrow 7.5=10-\frac{1000}{u^{2}} $
$\Rightarrow u^{2}=\frac{1000}{2.5}=400 $
or $ u=20\, m / s$
Hence, the initial speed of particle is $20\, m / s$.