Question

Two balls $A$ and $B$ are projected from the same height as shown in the figure. If the horizontal range of both the balls remains the same, then the speed $u$ of the second ball to the closest integer is $\dots \dots m{s}^{-1}.$

Answer 1

Answer: 11

The ball, $B$ , follows horizontal and angular projectile and the ball $A$ follows only horizontal projectile,
the height of the tower is, $h=490m$ , and both the particle follows the same range,
now for particle $A,$
$R=u\sqrt{\frac{2h}{g}}=10\times \sqrt{\frac{2\times 490}{9.8}}=100m$
and for oblique projectile,
$R=100m=ucos\theta \times t$
$\Rightarrow t=\frac{100}{ucos\left(\theta \right)}...\left(1\right)$
Now using it in vertical direction,
$\Rightarrow -490=usin\left(\theta \right)\times t-\frac{1}{2}g{t}^2$
$\Rightarrow -490=usin\left(30^{\circ }\right)\times \frac{100}{ucos\left(\theta \right)}-\frac{1}{2}g{\left(\frac{100}{ucos\left(\theta \right)}\right)}^2$
$\Rightarrow -490=usin\left(30^{\circ }\right)\times \frac{100}{ucos\left(30^{\circ }\right)}-\frac{1}{2}g{\left(\frac{100}{ucos\left(30^{\circ }\right)}\right)}^2$
$\Rightarrow -490=\frac{u}{2}\times \frac{2\times 100}{u\sqrt{3}}-\frac{1}{2}\times 9.8\times {\left(\frac{2\times 100}{u\sqrt{3}}\right)}^2$
$\Rightarrow -490=\frac{100}{\sqrt{3}}-\frac{196000}{u^2\times 3}$
$\Rightarrow \frac{196000}{u^2\times 3}=\frac{100}{\sqrt{3}}+490$
$\Rightarrow \frac{196000}{u^2\times 3}=\frac{100}{\sqrt{3}}+490$
$\Rightarrow u=10.9\approx 11m{s}^{-1}$