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A ball is thrown horizontally from the top of a building 25.3 m high. The ball strikes the ground at a point 32.9 m from the base of the building.The acceleration of gravity is 9.8 m/s2 . Find the time the ball is in motion. Answer in units of s

physics
Guest Sep 29, 2014

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 #2
avatar+26403 
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That is correct Chris!

There is also enough information to determine that the initial horizontal velocity is 32.9/2.27 ≈ 14.5 m/s.

Alan  Sep 29, 2014
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 #1
avatar+81029 
+10

I'll give this one a shot........

The equation for vertical the vertical motion of the projectile is given by:

y = (1/2)(-9.8m/s2)*t^2 + v0t       where y is the vertical displacement, t is the time and v0 is the initial vertical velocity (in this case = 0)    If we consider the top of the building to be  y = 0, then the vertical displacement = -25.3m

So we have

-25.3m = (-4.9m/s2)*t^2  divide both sides by  (-4.9m/s2)

(-25.3m) /  (-4.9m/s2) =  t^2   take the positive root of both sides

t = about 2.27 sec

I think this is correct....could you check it Alan or Melody ??

 

CPhill  Sep 29, 2014
 #2
avatar+26403 
+10
Best Answer

That is correct Chris!

There is also enough information to determine that the initial horizontal velocity is 32.9/2.27 ≈ 14.5 m/s.

Alan  Sep 29, 2014

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