Suppose that a ball is dropped from the upper observation de

Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground. What is the velocity of the ball after 2 seconds? How fast is the ball traveling when it hits the ground?

Solution

a)
s = ut +1/2 *at*t
u=0, since its dropped.
a=9.8
s=0*2+1/2 * 9.8 *2*2
s=19.6meters has it travelled for in 2 seconds.

b) When it hits the ground:

v^2 = u^0 +2as

v^2 = 0+2*9.8*450

v^2=8820
v==94m/sec
But the terminal velocity is 54m/sec
So, the velocity on hitting ground is 54m/sec