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Physics word problem. A ball is shot straight up from the ground at the initial velocity of 20m/s. 2 seconds later, a second ball is shot straight up from the ground. At what velocity must the second ball be shot at so that both balls land on the ground at the same time ?
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$$v_1t_1=v_2t_2\\20t_1=v_2(t_1-2)\\v_2=\frac{20t_1}{t_1-2}$$can you find out \(t_1\)?
hint: $$v_1t_1=-\frac12at_1^2$$where \(a=-9.8\ m/s^2\) for gravity
$$20t_1=-\frac12(-9.8)t_1^2\implies\dots $$
t1 = 4.08s
now use that in our first equation
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so v2 = 20 (4.08) / (4.08 - 2) = 39.23 m/s ?
Approximately, yes.
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