+0

# 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 bui

0
1083
2

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

### Best Answer

#2
+26619
+10

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
Sort:

### 2+0 Answers

#1
+85597
+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
+26619
+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

### 27 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details