(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.

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 04

Years : DEC 2014

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,

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

ii)

time = horizontal distance/horizontal speed

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