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.
- AmyLv 71 month ago
A ball accelerating at rate g takes 0.180 s to fall 1.90 m. Calculate its initial velocity.
A ball accelerating at rate g has initial velocity 0 and final velocity <Answer to part 1>. Calculate the distance traveled.
- oldschoolLv 71 month ago
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
- jeffrcalLv 71 month ago
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
- PhilomelLv 71 month ago
The distance =5.682m (with no air friction).
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- iblis666lovesyouLv 61 month ago
10 m above the window.
- oubaasLv 71 month ago
(Vi+(Vi+g*t))*t/2 = 1.90
3.80 = 2*Vi*0.18+9.806*0.18^2Vi = (3.80-9.806*0.18^2)/0.36 = 9.67 m/sec
h = Vi^2/2g = 9.67^2/19.612 = 4.77 m
- Andy CLv 71 month ago
I'm sorry. I can't ignore air Newtons aka resistance.