Question

The equation of trajectory of a projectile is given by $y=4x-\frac{40}{39}x^{2},$ where $y$ is vertically upwards. If $R$ is range and $\theta $ is angle of projection from horizontal, then value of $R$ $tan\theta $ is ( $x$ and $y$ are in $m$ )

• A. $15.6m$
• B. $3.9m$
• C. $11.7m$
• D. $14.2m$

Answer 1

Answer: (A)

$y=4x\left(1 - \frac{10 x}{39}\right)$
$\Rightarrow y=4x\left(1 - \frac{x}{3 . 9}\right)$
Equation of trajectory in terms of range is given by, $\Rightarrow y=xtan\left(\theta \right)\left(1 - \frac{x}{R}\right)$
On comparing with above equation, we have
$R=3.9mandtan\left(\theta \right)=4$
So $Rtan\left(\theta \right)=3.9\times 4=15.6m$