Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 2 seconds.

Answer:.

Required velocity = 6.26ms^-1

Explanation:

Given,

Distance, s = 450m

Time, t = 2 sec

Step 1. We obtain the distance covered within the given time under gravitational acceleration, g = 9.8ms^-2

S = ut + (1/2) gt^2. :; u = 0

: S = (1/2) gt^2

= (1/2) (9.8) (2^2)

= 19.6m

Step 2:

We obtain the velocity using the formula.

V^2 = u^2 + 2gs.

Where u is initial velocity, v is final / required velocity

Again u = 0

: V^2 = 2 (9.8) (19.6)

= 39.2

: V = 6.26ms^-1