Question

A ball is thrown horizontally from the top of a building 29.6 m high. The ball strikes the ground at a point 45 m from the base of the building. 
The acceleration of gravity is 9.8 m/s 
^2 
Find the y component of its velocity just before it strikes the ground.

Answer 1

Building Height h = 29.6 m

The acceleration of gravity is g = - 9.8 m/s²

Negative symbol of g indicates falling downwards direction of ball

Ball striking point from the base of the building x = 45 m

y component of its velocity = ?

From equation of motion : Distance s = s₀ + ut + ½at²

u = initial velocity

t = time

a = acceleration

From equation of motion : s = ut + ½at²

We are free to choose the origin of coordinate system wherever we want.Here choose the top of the building as origin ,so s = h = - 29.6 m

At vertical displacement

Substitute s = - 29.6 , a = g = - 9.8 , u = 0

- 29.6 = 0 + (¹/₂) ( - 9.8) t²

t² = 29.6×2/9.8

t = 2.46 sec

At horizontal displacement

From equation of motion : s = ut + ½at²

Substitute s = 45 , a = g = - 9.8

45 = u(2.46) + (¹/₂) ( - 9.8)(2.46)²

2.46u = 45 + 29.65

u = 30.34 m/s

At vertical( displacement

From equation of motion : v = u + at

Substitute u = 30.34 , a = g = - 9.8

v = u + at

v  = 30.34 + (-9.8)(2.46)

v  = - 24.108

v  = 6.24 m/s