Calculate the complete range of projection and the initial velocity of the projectile

Let velocity at A be $V_A$ makes an angle θ to the horizontal.

A to B

Horizontal motion:

Distance = velocity x time

$50=V_A \cos ⁡θ ×2.5 \\ V_A cos⁡θ=20 \space\space(1) $

Vertical motion,

$S=ut+\dfrac 12 at^2 \\ 0=V_A \sin ⁡θ ×2.5-1/2 ×9.81 ×2.5^2 \\ V_A \sin ⁡θ=\dfrac {0.5 ×9.81×2.5^2}{2.5} \\ V_A \sin θ =12.2625 \space\space (2) $

From (1) and (2)

$V_A=23.46 m/s \space \space And\space\space θ= 31.51^0 $

O To A

Horizontal motion:

$V_A \cos θ =u \cos ⁡α+ 0 \\ u \cos ⁡α=20\space\space (3) $

Vertical motion:

$V_y^2=u_y^2+ 2as \\ (V_A \sin θ)^2=(u \sin ⁡α )^2+ 2 ×(-9.81)×7.5 \\ (23.46 × \sin 31.51)^2+ 2 ×9.81 ×7.5=(u \sin ⁡α )^2 \\ u \sin ⁡α=17.2457 \space\space (4) $

From (3) and (4)

$\dfrac {u \sin ⁡α}{u \cos ⁡α } = \dfrac {17.2457}{20} \\ α= 40.77^0 $

From (3), $u=26.4086 m/s $

Range,

$R= \dfrac {u^2 \sin 2 α}g = \dfrac {26.4086^2 \sin (2 ×40.77)}{9.81 } \\ R=70.33 m $