A projectile is fired with a velocity v0 at an angle of 90- with an inclined plane having angle of inclination - with the horizontal- Vertical displacement of projectile when it strikes the incline for first time is

T=2v_0/gcosθ Y=v_0cosθ .T 1/2gT^2 =2v_0^2/g 2v_0^2/gcos^2θ=2v_0^2/gtan^2θ nbsp; ...

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Question

A projectile is fired with a velocity $v_{0}$ at an angle of 90° with an inclined plane (having angle of inclination $θ$ with the horizontal). Vertical displacement of projectile when it strikes the incline for first time is

\frac{2 v_{0}^{2}}{g} tan θ

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\frac{2 v_{0}^{2}}{g} sec θ

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\frac{2 v_{0}^{2}}{g} {sec}^{2} θ

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\frac{2 v_{0}^{2}}{g} {tan}^{2} θ

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Solution

The correct option is D $\frac{2 v_{0}^{2}}{g} {tan}^{2} θ$
$T = \frac{2 v_{0}}{g cos θ}$
$Y = v_{0} cos θ . T - \frac{1}{2} g T^{2}$
$= \frac{2 v_{0}^{2}}{g} - \frac{2 v_{0}^{2}}{g {cos}^{2} θ} = \frac{2 v_{0}^{2}}{g} {tan}^{2} θ$

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A projectile is fired with a velocity v0 at an angle of 90° with an inclined plane (having angle of inclination θ with the horizontal). Vertical displacement of projectile when it strikes the incline for first time is

The correct option is D 2v20gtan2θT=2v0gcosθ Y=v0cosθ.T−12gT2 =2v20g−2v20gcos2θ=2v20gtan2θ

A projectile is fired with a velocity $v_{0}$ at an angle of 90° with an inclined plane (having angle of inclination $θ$ with the horizontal). Vertical displacement of projectile when it strikes the incline for first time is

The correct option is D $\frac{2 v_{0}^{2}}{g} {tan}^{2} θ$
$T = \frac{2 v_{0}}{g cos θ}$
$Y = v_{0} cos θ . T - \frac{1}{2} g T^{2}$
$= \frac{2 v_{0}^{2}}{g} - \frac{2 v_{0}^{2}}{g {cos}^{2} θ} = \frac{2 v_{0}^{2}}{g} {tan}^{2} θ$