A ball is thrown upward from the ground with an initial speed of 16.6 m/s; at the same instant, another ball is dropped from a building 20 m high. After how long will the balls be at the same height above the ground?
Apply kinematic equation
the distance travelled first ball is
y1 -yo = u t - 1/2 at^2 ( yo = 0)
the distance travelled by the second ball is
y2-20 m = ut + 1/2 at^2 ( u= 0)
at y1= y2
u1t -1/2 at^2 = 20 m + 1/2 gt^2
16.6 m/s t - 1/2 * 9.8) t^2 = 20 m -1/2 (9.8) t^2
16.6 t = 20
t = 1.2 s
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