First slide

Projectile motion

Question

The angel made by relative motion of projectile 1 with respect to 2 with X- axis (horizontal) is:

Moderate

A

$θ = \tan^{- 1} (\frac{u_{12 y}}{u_{12 x}})$
B

$θ = \tan^{- 1} (\frac{u_{12 x}}{u_{12 y}})$
C

$θ = \sin^{- 1} (\frac{u_{12 x}}{u_{12 y}})$
D

$θ = \cos^{- 1} (\frac{u_{12 y}}{u_{12 x}})$

Solution

$p r o j e c t i l e 1 i n i t i a l v e l o c i t y = u_{1}; v e l o c i t y a t a n y p o i n t = v_{1} = u_{1 x} \hat{i} + (u_{1 y} - g t) \hat{j} p r o j e c t i l e 2 i n i t i a l v e l o c i t y = u_{2}; v e l o c i t y a t a n y p o i n t = v_{2} = i u_{2 x} + j (u_{2 y} - g t) r e l a t i v e v e l o c i t y o f p r o j e c t i l e 1 w i t h r e s p e c t t o 2 = v_{1} - v_{2} = i [u_{1 x} - u_{2 x}] + j [(u_{1 y} - g t) - (u_{2 y} - g t)] = i u_{12 x} + j u_{12 y} a n g e l m a d e b y r e l a t i v e m o t i o n o f p r o j e c t i l e 1 w i t h r e s p e c t t o θ, t h e n \tan θ = \frac{u_{12 y}}{u_{12 x}}$

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