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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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!
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