(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

\(\theta (t)=tan^{-1}(\dfrac{v_{0y-gt}}{v_{ox}})\)

(b) Shows that the projection angle θ0 for a projectile launched from the origin is given by

\(\theta_{0}=tan^{-1}(\dfrac{4h_{m}}{R})\)

where the symbols have their usual meaning

Asked by Abhisek | 1 year ago |  186

1 Answer

Solution :-

(a) Let θ be the angle at which the projectile is fired w.r.t the x-axis

θ depends on t

Therefore, tan θ(t) = \( \dfrac{v_x}{v_y}\) 

= \( \dfrac{ (v_{oy} – gt)}{v_{0x}}\)  (since vy = v0y -gt and vx =vox)

θ(t) = tan -1 \( \dfrac{ (v_{oy} – gt)}{v_{0x}}\)

 

(b) Since, hmax =\( \dfrac{ u^2 sin^2θ}{2g}\) —–(1)

R = \( \dfrac{ u^2 sin^2θ}{g}\)——–(2)

Dividing (1) by (2)

\( \dfrac{h_{max}}{R}\) =  \( [\dfrac{ u^2 sin^2θ}{2g}].[\dfrac{ u^2 sin^2θ}{2g}]\)

= tan \( \dfrac{ θ}{4}\)

θ = tan-1 (\( \dfrac{4h_{max}}{R}\))

Answered by Pragya Singh | 1 year ago

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