Easy Vertical Motion Question

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The discussion revolves around a vertical motion question regarding an object thrown at 18 m/s from a window, hitting the ground in 1.6 seconds. Participants debate whether the object was thrown upwards or downwards, noting that if thrown upwards, it would take longer than 1.6 seconds to hit the ground. Calculations show that if thrown downwards, the object would fall 41.3 meters, while if thrown upwards, it would indicate the ground is 16.3 meters above the window. Ultimately, the original question was deemed misleading, leading to its removal from the test. The clarification highlights the importance of precise wording in physics problems.
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"Easy" Vertical Motion Question

Hey,
This question was on my test today, but it didnt seem right to me..
Q) An object is thrown vertically at 18m/s from a window and hits the ground 1.6s later. What is the height of the window above the ground?


Is this even possible?
 
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Vertically up or down?

Use s = ut + 0.5 a*t^2

You got the t, you got the u, you got the a.

Seems pretty possible... :wink:
 
Vertically up.
But how can the entire trip only take 1.6s if the object is shot upwards at 18m/s? Wouldn't it take at least that long to slow the ball down so it could work on its down trip?
 
Were you specifically told "up"? If an object is thrown upwards at 18 m/s, then it will take 18/g= 18/9.8= 1.8 seconds for the object to reach its highest point. Obviously it can't hit the ground before that.

On the other hand, if the object is thrown vertically downwards at 18 m/s, the distance traveled at time t is x= 18t+ (g/2)t2.
With t= 1.6 seconds, g= 9.8 m/s2, that gives
x= 18(1.6)+ 4.9(1.6)2= 41.3 meters.

Another possibility is that the window is below ground!
Throwing an object up at 18 m/s gives a height of x= 18t- (g/2)t2. With t= 1.6 and g= 9.8, that gives x= 18(1.6)- (4.9)(1.6)2= 16.3 meters. The ground is 16.3 meters above the window!
 
Thanks for the help everyone!
It was up and I told my teacher and he took the question off the test...
 
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