From the top of a building 85.0 meters tall a ball is dropped. at the same time another ball is thrown upward from the ground with a speed of 46.0 m/s. a. how long after the balls are released will they hit? b. how high above the ground will these two balls hit?

a) Both balls are affected by gravity in the same way. The time both balls hit each other is equal to the time the ball thrown up reaches the top of the roof, if there was no gravity. (both balls would be at the same location)

y = v*t = 85m = 46m/s * t

t = 1.85s

Check:

The equation for position is given by:

y(t) = -0.5gt² + v₀t + x₀

The position of the ball on the roof top at time t = 1.85s:

y₁(t) = -4.9t² + 85.0 = 68.3m

The position of the ball thrown up at time t = 1.85s:

y₂(t) = -4.9t² + 46 * 1.85 = 68.3m

b) the two balls hit each other at a hight of 68.3m

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