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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
Solution 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)`
Solution 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`
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