Posted by **Chris** on Wednesday, August 25, 2010 at 5:06am.

Two bodies are thrown vertically upwards with the same initially velocity of 98metre/sec but 4 sec apart.How long after is thrown will they meet?

- Physics -
**drwls**, Wednesday, August 25, 2010 at 5:59am
In the United States,

"physical education" is the class devoted to exercise, fitness and sports. Your question is about physics.

Starting at t=0 when the first body is thrown, the height is

Y1 = 98t - 4.9 t^2. The second body's height, measured from the same t, is

Y2 = 98(t-4) - 4.9 (t-4)^2 (t>4)

Set the two equal and solve for t.

98t - 4.9 t^2 = 98(t-4) - 4.9 (t-4)^2

0 = -392 +4.9(8t)-16*4.9

39.2 t = 392 + 78.4 = 470.4

t = 12.0 seconds

The height at that time, for both objects, is 470.4 m

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