Easy Vertical Motion Question

"Easy" Vertical Motion Question

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?


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?


Science Advisor
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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