## anonymous one year ago Would any one derive and explain the formula... Range of a projectle=2 height of a projectile. Any help would be greatly appreciated...

1. anonymous

@irishboy123

2. anonymous

@arindameducationusc

3. anonymous

@michele_laino

4. IrishBoy123

can you scan or link the question pls :p

5. IrishBoy123

range R for generalised velocity v at angle $$\theta$$ is given by: $$R = \dfrac{v^2 \sin 2 \theta} {g}$$ the follows from equations of motion from $$v = u + at$$, we have half flight time $$t_{1/2}$$ is $$t_{1/2} = \dfrac{0 - v \sin \theta}{-g} = \dfrac{ v \sin \theta}{g}$$ thus horizontal range R is $$R = 2 \times t_{1/2} \times v \cos \theta = 2 \dfrac{ v \sin \theta}{g} v \cos \theta = \dfrac{v^2 \sin 2 \theta} {g}$$ the max height $$H_{max}$$ for that generalised trajectory comes from $$v^2 = u^2 + 2ax$$ and gives us $H_{max} = \dfrac{v^2 \sin^2 \theta}{2g}$ if you are looking for a trajectory where $$R = 2 H_{max}$$ then $$\dfrac{v^2 \sin 2 \theta} {g} = 2 \dfrac{v^2 \sin^2 \theta}{2g}$$ $$\implies \sin 2 \theta = \sin^2 \theta$$ $$\implies \tan \theta = 2$$

6. arindameducationusc

I agree with @IrishBoy123 The question must be like this During a projectile motion if the max height equals horizontal rangle, then the angle of projection with the horizontal is? And @IrishBoy123 has given the answer