(i) Horizontal distance AB it travel from the foot of the tower.

(ii) The velocity with which it strikes the ground at B.

(iii) Total time taken to reach point B

Given:

$U=20m/s \\ . α=30$

Let O be the origin.

Hence $0 = (0, 0)$

Hence, $B = (X,-50)$ (from the figure)

Hence, $x = ?; y = -50$

Using equation of path,

$Y= x \tan α – 4.9 x^2/ (u \cosα)^2 \\ -50 = X \tan30 – 4.9 \times X^2 /(20cos30)^2\\ X = 75.76 m$

ii)

time = horizontal distance/horizontal speed

$T= X/(u \cos α)= 75.76/(20 \cos30)= 4.37 sec$

iii)

$Vx=Ux = 20 \cos30→$ (because horizontal speed doesn’t change in a projectile motion)

$$Uy = 20 \sin30$$

$$Vy = ?$$

Using $v= u +at \\ Vy = Uy – g \times T \\ = 20\sin30 – 9.8 \times 4.37 \\ = -32.83 m/s = 32.83 m/s↓ \\ U = 37.12 m/s ↘ θ=62.2$