# A ball is dropped from an upper floor, some unknown distance above Tom's apartment.effects of air resistance. [...] physics help?

A ball is dropped from an upper floor, some unknown distance above Tom's apartment. As he looks out of his window, which is 1.90 m tall, Tom observes that it takes the ball 0.180 s to traverse the length of the window.

Determine how high ℎ above the top of Tom's window the ball was dropped. Ignore the effects of air resistance.

Relevance
• Part 1:

A ball accelerating at rate g takes 0.180 s to fall 1.90 m. Calculate its initial velocity.

Part 2:

A ball accelerating at rate g has initial velocity 0 and final velocity <Answer to part 1>. Calculate the distance traveled.

• Start out with d = V*t + ½*g*t²  where V is the speed at the top of the window and t is given to = 0.180s. Rearrange to find V

V = (d - ½*g*t²)/t =  (1.9 - 4.9*0.18²)/0.18 = 9.674m/s at the top of the window.

We know V² - Vi² = 2*g*d so (V² - Vi²)/(2*g) = d

(9.674² - 0²)/(2*9.8) = d = 4.77m above the top of the window

4.77m <<<<

• If we know how fast the ball is moving when it first appears in Tom's window, this will indicate the height from which it is dropped. We can find that value thusly:

D = .5gt^2 +vt

substituting and solving for v:

v =[1.90m - .5(9.8)(.180)^2]/.180 = 9.67m/s

So for a ball accelerating under gravity to reach 9.67m/s requires a time of:

v = at => t=9.67m/s/9.8m/s^2 = .987 seconds

Using that figure to solve for h:

h= .5*9.8*.987^2 = 4.77m

• 1.9/.18=10.556m/s

The distance =5.682m (with no air friction).