Answer to Question #165020 in Mechanics | Relativity for gabriel

Question

a particle is projected with velocity whose horizontal and vertical component are U and V respectively and the particle passes through a point of projection are H and K respectively

prove 2U^2K+gH=2UVH

Answer 1

Given Parameters-

Let the velocity of projection be $v$ , and

angle of projection be $\theta$

$H=\text{horizontal component of velocity}=vcos\theta$

$V=\text{Vertical component of velocity}=vsin\theta$

$tan\theta=\dfrac{vsin\theta}{vcos\theta}=\dfrac{V}{U}~~~~~-(1)$

The trajectory Equation for projectile motion is given by-

$y=\text{x tan}\theta-\dfrac{gx^2}{2v^2cos^2\theta}$

Since The particle Passes through $(H,K)$

$\Rightarrow K=Htan\theta-\dfrac{gH^2}{2(vcos\theta)^2}$

Substitute the above Parameter in above equation:-

$\Rightarrow K=\dfrac{HV}{U}-\dfrac{gH^2}{2U^2}$

$\Rightarrow K=\dfrac{2UHV-gH^2}{2U^2}$

$\Rightarrow 2U^2K=2UHV-gH^2$

$\Rightarrow 2U^2K+gH^2=2UHV$

Hence proved.