Question

Question #dc83d

Answer 1

`E_p/E_k=sec^2 alpha-1`

Explanation:

`v=v_x=v_i*cos alpha" (Horizontal component of velocity of object)"`

`E_k=1/2*m*v_x^2`

`E_k=1/2*m*v_i^2*cos^2 alpha`

`h_m=v_i^2*sin^2 alpha/(2*g)" maximum height of object"`

`E_p=m*g*h_m`

`E_p=m*cancel(g)*(v_i^2*sin^2 alpha)/(2*cancel(g))`

`E_p=(m v_i^2*sin^2 alpha)/2`

`E_p/E_k=(cancel(1/2*m)*cancel(v_i^2)*sin^2 alpha)/(cancel(1/2*m) cancel(v_i)^2*cos^2alpha)`

`E_p/E_k=sin^2 alpha/cos^2 alpha`

`sin^2 alpha=1-cos^2 alpha`

`E_p/E_k=(1-cos^2 alpha)/(cos^2 alpha)`

`E_p/E_k=(1/cos^2 alpha)-1`

`1/cos^2 alpha=sec^2 alpha`

`E_p/E_k=sec^2 alpha-1`

Answer 2

As referred to the figure above.

Maximum height of the object having mass `m` will be reached when `y`- or `sin theta` component of projectile's velocity becomes `0` (zero).

As such kinetic energy at this point is given by
`KE_"max height"=1/2m(ucos theta)^2`

At this point all the kinetic energy due to `y`-component of initial velocity has got converted into its Potential Energy which is also given as `mgh`
`PE_"max height"=mgh=1/2m(usin theta)^2`

Ratio of the two is `(PE_"max height")/(KE_"max height")=(1/2m u^2sin^2 theta)/(1/2m u^2cos^2theta)`
`=tan^2 theta`