Question

Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450m above the ground.
a) What is the velocity of the ball after 5 seconds?
B) How fast is the ball when it hits the ground?

Thanks :)

Answer 1

For this problem, I'm converting meters to feet as I am using -32 ft/sec^2 as the gravitational constant, so:

1 Meter = 3.2808399 Feet

450 m = 1476.377955 feet

The position function, using the standard gravitational constant at sea level, is:

s(t) = -16t^2 + (V0)t + (S0)

and the velocity function is:

v(t) = -32t + (V0)

At t=5, you have for velocity:

v(5) = -32(5) + (0) V0 is 0 because the ball is dropped

For the ball hitting the ground:

s(t) = 0 = -16t^2 + (V0)t + 1476.377955

0 = -16t^2 + (0)t + 1476.377955

0 = -16t^2 + 1476.377955 -> t = sqrt(1476.377955/16) = 9.6059

v(9.6059) = -32(9.6059)

Answer 2

can velocity be negative ??

Answer 3

Sure! Velocity is a vector, it has magnitude and direction. Speed has magnitude only and is expressed always as positive.

Answer 4

Speed = |velocity|

Answer 5

I did this problem in feet per sec instead of meters per second. Is that ok for you? If not, then you have to re-convert the velocities back to meters/sec. Can you do that?

Answer 6

yes that is okay :) Thank you

Answer 7

uw! Good luck to you in all of your studies and thx for the recognition! @joselin12