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Standard XII

Physics

Work Done as Dot Product

Derive the fo...

Question

Derive the formula for the range of a projectile thrown with a velocity $u$ at an angle $θ$ from the horizontal.

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Solution

Here, $u_{x} = u c o s θ$
and $u_{y} = u s i n θ$
At maximum height, $v_{y} = 0 \Rightarrow u s i n θ - g t = 0 \Rightarrow t = \frac{u s i n θ}{g}$
So, time of flight (T) $= 2 t = \frac{2 u s i n θ}{g}$
and horizontal range $= u c o s θ \times T = \frac{u^{2} 2 s i n θ c o s θ}{g} = \frac{u^{2} s i n 2 θ}{g}$

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Derive the formula for the range of a projectile thrown with a velocity u at an angle θ from the horizontal.

Here, ux=ucosθ and uy=usinθ At maximum height, vy=0⇒usinθ−gt=0⇒t=usinθg So, time of flight (T) =2t=2usinθg and horizontal range =ucosθ×T=u22sinθcosθg=u2sin2θg

Derive the formula for the range of a projectile thrown with a velocity $u$ at an angle $θ$ from the horizontal.