Question

Show that the projection angle `theta_o` for a projectile launched from the origin is given by

`theta_o =tan^(-1) ((4h_m)/R)`

Where the symbols have their usual meaning

Answer 1

Maximum vertical height `h_m = (v_o^2 sin^2theta)/2g` ...(i)

Horiaontal range `R = (v_o^2 sin 2theta)/g` .. (ii)

Solving equations (i) and (ii), we get:

`h_m/R = (sin^2 theta)/(2sin 2theta) = (sin^2theta)/(4sin theta cos theta) = sin theta/(4cos theta)`

`=>h_m/R = (Tan theta)/4`

`=> theta = Tan^(-1) ((4h_m)/R)`

Answer 2

Since h_"max" = `(u^2sin^2theta)"/2g"`

and `R = (u^2sin 2theta)/g`

`=> h_"max"/R = (u^2sin^2theta"/2g")/(u^2sin 2theta"/g") = (tan theta)/4`

`=> (tan theta)/4 = h_"max"/R`

or `tan theta = (4h_"max")/R`

or `theta = tan^(-1)(4h_"max")/R`