Easy Vertical Motion Question

[Inquiring_Mike]

"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?

 

[FZ+]

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...

 

[Inquiring_Mike]

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?

 

[HallsofIvy]

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!

 

[Inquiring_Mike]

Thanks for the help everyone!
It was up and I told my teacher and he took the question off the test...