The equation of trajectory of a projectile is given by y=4x-(40/39)x2, where y is vertically upwards. If R is range and θ is angle of projection from horizontal, then value of R tanθ is ( x and y are in m )

Question

The equation of trajectory of a projectile is given by y=4x-(40/39)x2, where y is vertically upwards. If R is range and θ is angle of projection from horizontal, then value of R tanθ is ( x and y are in m ) (A) 15.6m (B) 3.9m (C) 11.7m (D) 14.2m

Tardigrade Admin · Accepted Answer

y=4x(1 - (10 x/39)) ⇒ y=4x(1 - (x/3 . 9)) Equation of trajectory in terms of range is given by, ⇒ y=xtan(θ )(1 - (x/R)) On comparing with above equation, we have R=3.9mandtan(θ )=4 So Rtan(θ )=3.9× 4=15.6m

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

$15.6 m$

$3.9 m$

$11.7 m$

$14.2 m$

$y = 4 x (1 - \frac{10 x}{39})$
$\Rightarrow y = 4 x (1 - \frac{x}{3.9})$
Equation of trajectory in terms of range is given by, $\Rightarrow y = x t an (θ) (1 - \frac{x}{R})$
On comparing with above equation, we have
$R = 3.9 man d t an (θ) = 4$
So $Rt an (θ) = 3.9 \times 4 = 15.6 m$

Q. The equation of trajectory of a projectile is given by y=4x−3940​x2, where y is vertically upwards. If R is range and θ is angle of projection from horizontal, then value of R tanθ is ( x and y are in m )

15.6m

3.9m

11.7m

14.2m