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(t4)  4.9 (t4)^2 (t>4)
Set the two equal and solve for t.
98t  4.9 t^2 = 98(t4)  4.9 (t4)^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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