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 $7.5m$ above the ground while it travels a horizontal distance of $10m$ from the point of projection, then the initial speed of particle is
(Assume $g=10m/s^2$ )

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

Answer 1

Answer: (B)

The given situation is as shown in figure,

Let the particle reaches point $P$ in time $t$.
Then, horizontal distance, $x=u\cos \theta t$
$\Rightarrow 10=u\cos 45^{\circ }\times t$
$\Rightarrow t=\frac{10\sqrt{2}}{u}\dots$ (i)
and vertical distance,
$y=u\sin \theta t-\frac{1}{2}g{t}^2$
$\Rightarrow 7.5=u\sin 45^{\circ }\times t-\frac{1}{2}\times 10\times t^2\dots$ (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 $u=20m/s$
Hence, the initial speed of particle is $20m/s$.