# A red ball is thrown downward with an initial speed of 10 m/s at time t = 0 from a vertical height of 80 m above the ground. At time t = 2 s later, a green ball is thrown vertically upward from the ground with an initial speed of 30 m/s. At what height above the ground do these objects pass each other, if possible?

Mar 7, 2014

The objects pass each other at a height of 18 m.

Choose ground level as 0, so “up” is positive and “down” is negative. Let $g$ = acceleration due to gravity, $t$ = time, ${v}_{0}$ = initial velocity, ${y}_{0}$ = the initial height, $y$ = height at time $t$. Then

y = y_0 + v_0t + ½g t^2

Height of red ball =
y_(red) = 80 - 10t – 4.905t^2

Since the blue ball starts 2 s later, height of blue ball =
${y}_{b l u e} = 0 + 30 \left(t - 2\right) - 4.90 {\left(t - 2\right)}^{2}$ =

30t – 60 – 4.90(t^2-4t+4) = 30t – 60 – 4.90t^2 + 19.6t - 19.6 =

-79.6 + 49.6t – 4.90t^2

When they meet, ${y}_{red} = {y}_{b l u e}$.
80 - 10t – 4.90t^2 = -79.6 + 49.6t – 4.90t^2
$80 - 10 t = - 79.6 + 49.6 t$
$60 t = 160$
$t = \frac{160}{60}$ = 2.7 s

At 2.7 s

y_(red) = 80 - 10t – 4.90t^2 = 80 – 10×2.7 – 4.90×2.7^2 =

80 – 27 – 35 = 18 m

y_(blue) = 0 + 30(t-2) - 4.90(t-2)^2 = 30× 0.7 – 4.90×0.7^2 =

20 – 2 = 18 m

The objects pass each other at a height of 18 m.

Note: I have calculated the answer to only two significant figures, because that is all the question gave for the height and the speeds. If you need more precision, you will have to recalculate.