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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.

User JeanT
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1 Answer

4 votes
4 votes

Answer:.

Required velocity = 6.26ms^-1

Step-by-step 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

User Banty Roy
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