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

Guest Sep 29, 2014

#1**+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/s^{2})*t^2 + v_{0}t where y is the vertical displacement, t is the time and v_{0} 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/s^{2})*t^2 divide both sides by (-4.9m/s^{2})

(-25.3m) / (-4.9m/s^{2}) = 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