Question #250f9

Apr 1, 2016

Four times.

Explanation:

Let's assume that second body is thrown with a velocity $\vec{u}$ making an angle $\theta$ with $x$ axis.

Velocity along $x$-axis $= {u}_{x} = u \cos \theta$
Acceleration along $x$-axis ${a}_{x} = 0$
Velocity along $y$-axis $= {u}_{y} = u \sin \theta$
Acceleration along $y$-axis ${a}_{y} = - g$

To obtain total Time of flight $t$, only $y$ component of velocity needs to be considered. The ball rises up, comes to a stop and then starts its decent downwards. Once it reaches the ground level, magnitude of the $y$ component of velocity is same when it was thrown. Only the direction is opposite. Using the equation

$v = u + a t$, and inserting given values
$- u \sin \theta = u \sin \theta - g t$

$t = 2 u \sin \frac{\theta}{g}$.........(1)
Horizontal Range$= \text{Horizontal velocity" xx "Time of flight}$
$R = u \cos \theta \times 2 u \sin \frac{\theta}{g}$
$= {u}^{2} / g 2 \sin \theta \cos \theta$
$= \frac{{u}^{2} \sin 2 \theta}{g}$ ....(2)

It is given that velocity of first body is double that of the second body and angle $\theta$ is same
From (2) we see that for the same angle of throw Range $R \propto {u}^{2}$.
$\implies$ Distance covered by first body is ${2}^{2} = 4$ times the range of second.