(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

##### 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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