posted by Chris on .
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?
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