A projectile is fired with a velocity v at right angle to the slope which is inclined at an angle θ with the horizontal. The range of the projectile along the inclined plane is:

Question

A projectile is fired with a velocity v at right angle to the slope which is inclined at an angle θ with the horizontal. The range of the projectile along the inclined plane is: (A) (2 v2 tan θ/g) (B) (v2 sec θ/g) (C) 2 v2 tan θ sec θ (D) (v2 sin θ/g)

Tardigrade Admin · Accepted Answer

If t is the time of flight, then 0=v t-(1/2)(g cos θ) t2 ⇒ t=(2 v/g) g cos θ R=0+(1/2)(g sin θ) t2=(1/2) g sin θ((2 v/g cos θ))2 =(2 v2/g) tan θ sec θ

Q. A projectile is fired with a velocity $v$ at right angle to the slope which is inclined at an angle $θ$ with the horizontal. The range of the projectile along the inclined plane is:

$\frac{2 v ^{2} t a n θ}{g}$

$\frac{v ^{2} s e c θ}{g}$

$2 v^{2} tan θ sec θ$

$\frac{v ^{2} s i n θ}{g}$

If $t$ is the time of flight, then
$0 = v t - \frac{1}{2} (g cos θ) t^{2}$
$\Rightarrow t = \frac{2 v}{g} g cos θ$
$R = 0 + \frac{1}{2} (g sin θ) t^{2} = \frac{1}{2} g sin θ (\frac{2 v}{g c o s θ})^{2}$
$= \frac{2 v ^{2}}{g} tan θ sec θ$

Q. A projectile is fired with a velocity v at right angle to the slope which is inclined at an angle θ with the horizontal. The range of the projectile along the inclined plane is:

g2v2tanθ​

gv2secθ​

2v2tanθsecθ

gv2sinθ​