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
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 ??